This document discusses simple harmonic motion. It begins by introducing simple harmonic motion as the to-and-fro motion of a particle vibrating about a central mean position. It then provides the differential equation that describes simple harmonic motion and discusses key terms like amplitude, period, and frequency. The document also examines simple harmonic motion in pendulums, springs, and other contexts, establishing various governing laws and relationships.
This document discusses beam theory and provides equations for calculating the deflection and slope of beams under different loading conditions. It defines key terms like bending moment, radius of curvature, flexural stiffness, and provides equations relating these terms. Specifically, it gives the deflection and slope equations for a cantilever beam with a point load, cantilever with uniform load, simply supported beam with central point load, and simply supported beam with uniform load.
This document discusses beam design criteria and deflection behavior of beams. It covers two key criteria for beam design:
1) Strength criterion - the beam cross section must be strong enough to resist bending moments and shear forces.
2) Stiffness criterion - the maximum deflection of the beam cannot exceed a limit and the beam must be stiff enough to resist deflections from loading.
It then defines deflection, slope, elastic curve, and flexural rigidity. It presents the differential equation that relates bending moment, slope, and deflection. Methods for calculating slope and deflection including double integration, Macaulay's method, and others are also summarized.
This document provides information about shear stresses and shear force in structures. It includes:
- Definitions of shear force and shear stress. Shear force is an unbalanced force parallel to a cross-section, and shear stress develops to resist the shear force.
- Explanations of horizontal and vertical shear stresses that develop in beams due to bending moments. Shear stress is highest at the neutral axis and reduces towards the top and bottom of the beam cross-section.
- Derivations of formulas for calculating shear stress across different beam cross-sections. Shear stress is directly proportional to the shear force and beam geometry.
- Examples of calculating maximum and average shear stresses for various cross-sections
The document discusses methods for calculating deflections in structures, specifically the moment area method. It provides examples of using the moment area method to calculate slopes and deflections at various points along beams and frames by relating the bending moment diagram area to slope changes and vertical deflections using theorems. Sample problems are worked through step-by-step to demonstrate calculating slopes and deflections for beams under different loading conditions.
2-vector operation and force analysis.pptRanaUmair74
This document discusses vector operations and force analysis. It begins with an overview of key concepts related to vectors, including defining scalars and vectors, and methods for finding the resultant force of multiple vectors using graphical and analytical approaches. It then covers topics such as resolving forces into rectangular components, adding vectors, and determining the magnitude and direction of resultant forces. Examples are provided to demonstrate how to apply these techniques to solve force analysis problems involving both 2D and 3D systems of forces.
12. Design of Machine Elements -Belt drive.pptxPraveen Kumar
This document discusses the design of belt drives. It begins with an introduction to belt drives, their advantages and disadvantages. It then discusses belt drive selection criteria and types of belts and materials. The document covers various types of flat belt drives including open, crossed, and compound drives. It derives relationships for velocity ratio and belt tensions. It also addresses slip and creep in belts and how they affect power transmission efficiency. The summary covers the key aspects and applications of belt drive design.
This document discusses beam theory and provides equations for calculating the deflection and slope of beams under different loading conditions. It defines key terms like bending moment, radius of curvature, flexural stiffness, and provides equations relating these terms. Specifically, it gives the deflection and slope equations for a cantilever beam with a point load, cantilever with uniform load, simply supported beam with central point load, and simply supported beam with uniform load.
This document discusses beam design criteria and deflection behavior of beams. It covers two key criteria for beam design:
1) Strength criterion - the beam cross section must be strong enough to resist bending moments and shear forces.
2) Stiffness criterion - the maximum deflection of the beam cannot exceed a limit and the beam must be stiff enough to resist deflections from loading.
It then defines deflection, slope, elastic curve, and flexural rigidity. It presents the differential equation that relates bending moment, slope, and deflection. Methods for calculating slope and deflection including double integration, Macaulay's method, and others are also summarized.
This document provides information about shear stresses and shear force in structures. It includes:
- Definitions of shear force and shear stress. Shear force is an unbalanced force parallel to a cross-section, and shear stress develops to resist the shear force.
- Explanations of horizontal and vertical shear stresses that develop in beams due to bending moments. Shear stress is highest at the neutral axis and reduces towards the top and bottom of the beam cross-section.
- Derivations of formulas for calculating shear stress across different beam cross-sections. Shear stress is directly proportional to the shear force and beam geometry.
- Examples of calculating maximum and average shear stresses for various cross-sections
The document discusses methods for calculating deflections in structures, specifically the moment area method. It provides examples of using the moment area method to calculate slopes and deflections at various points along beams and frames by relating the bending moment diagram area to slope changes and vertical deflections using theorems. Sample problems are worked through step-by-step to demonstrate calculating slopes and deflections for beams under different loading conditions.
2-vector operation and force analysis.pptRanaUmair74
This document discusses vector operations and force analysis. It begins with an overview of key concepts related to vectors, including defining scalars and vectors, and methods for finding the resultant force of multiple vectors using graphical and analytical approaches. It then covers topics such as resolving forces into rectangular components, adding vectors, and determining the magnitude and direction of resultant forces. Examples are provided to demonstrate how to apply these techniques to solve force analysis problems involving both 2D and 3D systems of forces.
12. Design of Machine Elements -Belt drive.pptxPraveen Kumar
This document discusses the design of belt drives. It begins with an introduction to belt drives, their advantages and disadvantages. It then discusses belt drive selection criteria and types of belts and materials. The document covers various types of flat belt drives including open, crossed, and compound drives. It derives relationships for velocity ratio and belt tensions. It also addresses slip and creep in belts and how they affect power transmission efficiency. The summary covers the key aspects and applications of belt drive design.
This document discusses friction and provides an example problem. It defines friction as the force resisting the relative motion of surfaces in contact. There are two types of friction: static and kinetic. The key elements of friction are defined, including the coefficient of friction μ and angle of friction φ. Laws of friction relate φ to the angle of an applied force θ. An example problem then applies these concepts to calculate the minimum force P required to start motion of a 400 lb block on a rough surface with a coefficient of friction of 0.40.
Here are the key steps to solve this problem:
1) Calculate the centrifugal forces at the minimum and maximum radii using Fc = mω2r
2) Use the lever equation to relate the centrifugal forces to the spring forces:
Fc1/S1 = r1/x
Fc2/S2 = r2/x
3) The initial compression of the spring is r2 - r1
4) Use the relation between spring force and compression to find the spring constant:
ΔS/Δx = s
Where ΔS is the change in spring force (S2 - S1) and Δx is the change in compression (r2 - r1
Fluid Mechanics Chapter 3. Integral relations for a control volumeAddisu Dagne Zegeye
Introduction, physical laws of fluid mechanics, the Reynolds transport theorem, Conservation of mass equation, Linear momentum equation, Angular momentum equation, Energy equation, Bernoulli equation
This document discusses the shear center of beam sections. It defines the shear center as the point where a load can be applied to cause pure bending without any twisting. It then provides properties of the shear center, including that it lies on the axis of symmetry for some sections. Methods for determining the location of the shear center are presented, including using the first moment of area. Real-life examples of applying shear center concepts to purlins and channel sections are given. The document concludes with an example problem of locating the shear center and calculating shear stresses for a hat section.
The document discusses shear force and bending moment diagrams. It defines shear force and bending moment, explaining that shear force acts perpendicular to the beam's axis while bending moment acts to bend the beam. It outlines the procedure to determine shear force and bending moment diagrams: (1) calculate support reactions, (2) divide the beam into segments based on loading, (3) draw free body diagrams and calculate expressions for each segment. As an example, it analyzes a simply supported beam with two loads to derive the shear force and bending moment expressions and diagrams.
1) To analyze accelerations, positions must first be found to calculate velocities by differentiation and accelerations by further differentiation.
2) Acceleration has two components - tangential and centripetal. For uniform motion only centripetal acceleration exists, and for straight-line motion only tangential acceleration exists.
3) The Coriolis component arises for points moving on rotating links and is perpendicular to the link and proportional to the product of linear and angular velocities.
The document discusses shear stresses in beams. It defines shear stress as being due to shear force and perpendicular to the cross-sectional area. Shear stress is derived as τ = F/A, where F is the shear force and A is the cross-sectional area. Shear stress varies across standard beam cross sections like rectangular, circular, and triangular. Shear stress is maximum at the neutral axis for rectangular and circular beams, and at half the depth for triangular beams. Sample problems are included to demonstrate calculating and graphing the variation of shear stress across specific beam cross sections.
Pure bending of curved bar (polar coordinate)Pratish Sardar
This document discusses pure bending of curved bars. It presents the assumptions made in analyzing curved members under bending, including that plane cross sections remain plane after bending. It defines key parameters like the radii of the inner and outer fibers. The stress function is introduced and the equations for radial and tangential stresses are given. The boundary conditions for a curved bar under pure bending are described. Standard relations are presented for determining the coefficients of the stress function based on the boundary conditions. Radial and circumferential stress equations are provided in terms of these coefficients. Finally, the document indicates it will provide numerical examples.
1) Lateral loads on a beam cause it to bend into a deflection curve. Pure bending occurs when the bending moment is constant, resulting in zero shear force. Non-uniform bending happens when the bending moment is variable, with non-zero shear force.
2) The radius of curvature and curvature of a beam are defined based on the deflection curve. Longitudinal strains in a beam vary linearly with distance from the neutral axis, causing elongation on one side and shortening on the other.
3) For a simply supported beam bent into a circular arc, the example calculates the radius of curvature, curvature, and midpoint deflection based on the given bottom strain. The normal stress is directly proportional to
Ekeeda Provides Online Civil Engineering Degree Subjects Courses, Video Lectures for All Engineering Universities. Video Tutorials Covers Subjects of Mechanical Engineering Degree.
This document discusses friction in pivot and collar bearings. It begins by providing examples of rotating shafts that experience axial thrust, such as propeller shafts and turbine shafts. It then discusses assumptions made in studying friction in bearings, including uniform pressure and wear distributions. Several pages are dedicated to analyzing the friction torque in flat pivot bearings assuming uniform pressure and uniform wear. A similar analysis is done for collar bearings. Formulas are provided for calculating friction torque and power lost to friction in various bearing configurations. Examples are included to demonstrate calculations.
This document discusses the kinematics of particles in rectilinear and curvilinear motion. It defines key concepts like position, displacement, velocity, and acceleration for both continuous and erratic rectilinear motion. Examples are provided to demonstrate how to construct velocity-time and acceleration-time graphs from a given position-time graph, and vice versa. The chapter then discusses general curvilinear motion, defining position, displacement, velocity, and acceleration using vector analysis since the curved path is three-dimensional. Fundamental problems and practice problems are also included.
This document provides information about the Strength of Materials CIE 102 course for first year B.E. degree students. It includes a list of 10 topics that will be covered in the course, such as simple stress and strain, shearing force and bending moment, and stability of columns. It also lists several reference books for the course and provides an overview of concepts that will be discussed in the first chapter, including stress, strain, stress-strain diagrams, and ductile vs brittle materials.
Any mechanical system can be classified by the number of degrees of freedom (DOF) it possesses, which is equal to the number of independent parameters needed to uniquely define its position in space at any given time. A DOF is defined relative to a selected frame of reference. The document discusses determining a mechanism's DOF using Kutzback's equation and provides examples of mechanisms with one and zero DOF that contain different types of joints.
The document discusses various topics related to stress and strain including: principal stresses and strains, Mohr's stress circle theory of failure, 3D stress and strain, equilibrium equations, and impact loading. It provides examples of stresses acting in different planes including normal, shear, oblique, and principal planes. It also gives examples of calculating normal and tangential stresses on an oblique plane subjected to stresses in one, two, or multiple directions with and without shear stresses.
The document discusses compound stresses, which involve both normal and shear stresses acting on a plane. It provides equations to calculate:
1) Normal and shear stresses on a plane inclined to the given stress plane.
2) The inclination and normal stresses on the planes of maximum and minimum normal stress (principal planes).
3) The inclination and shear stresses on the planes of maximum shear stress.
It includes an example problem calculating the principal stresses and maximum shear stresses given a state of stress. Sign conventions for stresses are also defined.
Overhanged Beam and Cantilever beam problemssushma chinta
This document discusses shear force diagrams (SFD) and bending moment diagrams (BMD) for overhanging beams and cantilever beams under different loading conditions. It provides examples of overhanging beams with uniform distributed loading and analyses the reactions, shear forces, bending moments, and point of contraflexure. It also discusses cantilever beams and provides examples of cantilevers with point loads and uniform distributed loads, deriving the corresponding SFDs and BMDs.
The document discusses various types of frictional clutches, including disc clutches, cone clutches, and centrifugal clutches. It provides details on the structure and operating principles of single plate and multi-plate disc clutches. The torque transmission analysis for these clutches is presented based on both uniform pressure and uniform wear theories. Cone clutches are described as providing higher torque transfer compared to disc clutches of the same size. Centrifugal clutches are explained as using centrifugal force to automatically engage and disengage based on engine speed.
Torsion (Springs)-Strength of Materials.pdfDr.S.SURESH
This document discusses different types of springs, including helical and leaf springs. Helical springs are made of coiled wire and store energy through torsion or twisting. Leaf springs are laminated and absorb shocks in vehicles. The document provides equations to calculate spring deflection, stiffness, and maximum shear/bending stress. It includes examples solving for the number of plates in a leaf spring, length of a leaf spring, and deflection under a given load.
1) Simple harmonic motion is the motion of an object where the acceleration is directly proportional to the displacement from the equilibrium position and directed towards the equilibrium.
2) It can be modeled as circular motion where the acceleration towards the center is proportional to the displacement from the center.
3) Simple harmonic oscillators include spring-mass systems and pendulums, where the restoring force is proportional to the displacement.
1) Simple harmonic motion (SHM) is a type of periodic motion where an object moves back and forth over the same path, like a mass on a spring or a pendulum.
2) For motion to be SHM, there must be a restoring force acting towards the equilibrium position that is proportional to the displacement.
3) The acceleration during SHM is directly proportional to the displacement from the equilibrium position and always acts to restore the object towards equilibrium.
This document discusses friction and provides an example problem. It defines friction as the force resisting the relative motion of surfaces in contact. There are two types of friction: static and kinetic. The key elements of friction are defined, including the coefficient of friction μ and angle of friction φ. Laws of friction relate φ to the angle of an applied force θ. An example problem then applies these concepts to calculate the minimum force P required to start motion of a 400 lb block on a rough surface with a coefficient of friction of 0.40.
Here are the key steps to solve this problem:
1) Calculate the centrifugal forces at the minimum and maximum radii using Fc = mω2r
2) Use the lever equation to relate the centrifugal forces to the spring forces:
Fc1/S1 = r1/x
Fc2/S2 = r2/x
3) The initial compression of the spring is r2 - r1
4) Use the relation between spring force and compression to find the spring constant:
ΔS/Δx = s
Where ΔS is the change in spring force (S2 - S1) and Δx is the change in compression (r2 - r1
Fluid Mechanics Chapter 3. Integral relations for a control volumeAddisu Dagne Zegeye
Introduction, physical laws of fluid mechanics, the Reynolds transport theorem, Conservation of mass equation, Linear momentum equation, Angular momentum equation, Energy equation, Bernoulli equation
This document discusses the shear center of beam sections. It defines the shear center as the point where a load can be applied to cause pure bending without any twisting. It then provides properties of the shear center, including that it lies on the axis of symmetry for some sections. Methods for determining the location of the shear center are presented, including using the first moment of area. Real-life examples of applying shear center concepts to purlins and channel sections are given. The document concludes with an example problem of locating the shear center and calculating shear stresses for a hat section.
The document discusses shear force and bending moment diagrams. It defines shear force and bending moment, explaining that shear force acts perpendicular to the beam's axis while bending moment acts to bend the beam. It outlines the procedure to determine shear force and bending moment diagrams: (1) calculate support reactions, (2) divide the beam into segments based on loading, (3) draw free body diagrams and calculate expressions for each segment. As an example, it analyzes a simply supported beam with two loads to derive the shear force and bending moment expressions and diagrams.
1) To analyze accelerations, positions must first be found to calculate velocities by differentiation and accelerations by further differentiation.
2) Acceleration has two components - tangential and centripetal. For uniform motion only centripetal acceleration exists, and for straight-line motion only tangential acceleration exists.
3) The Coriolis component arises for points moving on rotating links and is perpendicular to the link and proportional to the product of linear and angular velocities.
The document discusses shear stresses in beams. It defines shear stress as being due to shear force and perpendicular to the cross-sectional area. Shear stress is derived as τ = F/A, where F is the shear force and A is the cross-sectional area. Shear stress varies across standard beam cross sections like rectangular, circular, and triangular. Shear stress is maximum at the neutral axis for rectangular and circular beams, and at half the depth for triangular beams. Sample problems are included to demonstrate calculating and graphing the variation of shear stress across specific beam cross sections.
Pure bending of curved bar (polar coordinate)Pratish Sardar
This document discusses pure bending of curved bars. It presents the assumptions made in analyzing curved members under bending, including that plane cross sections remain plane after bending. It defines key parameters like the radii of the inner and outer fibers. The stress function is introduced and the equations for radial and tangential stresses are given. The boundary conditions for a curved bar under pure bending are described. Standard relations are presented for determining the coefficients of the stress function based on the boundary conditions. Radial and circumferential stress equations are provided in terms of these coefficients. Finally, the document indicates it will provide numerical examples.
1) Lateral loads on a beam cause it to bend into a deflection curve. Pure bending occurs when the bending moment is constant, resulting in zero shear force. Non-uniform bending happens when the bending moment is variable, with non-zero shear force.
2) The radius of curvature and curvature of a beam are defined based on the deflection curve. Longitudinal strains in a beam vary linearly with distance from the neutral axis, causing elongation on one side and shortening on the other.
3) For a simply supported beam bent into a circular arc, the example calculates the radius of curvature, curvature, and midpoint deflection based on the given bottom strain. The normal stress is directly proportional to
Ekeeda Provides Online Civil Engineering Degree Subjects Courses, Video Lectures for All Engineering Universities. Video Tutorials Covers Subjects of Mechanical Engineering Degree.
This document discusses friction in pivot and collar bearings. It begins by providing examples of rotating shafts that experience axial thrust, such as propeller shafts and turbine shafts. It then discusses assumptions made in studying friction in bearings, including uniform pressure and wear distributions. Several pages are dedicated to analyzing the friction torque in flat pivot bearings assuming uniform pressure and uniform wear. A similar analysis is done for collar bearings. Formulas are provided for calculating friction torque and power lost to friction in various bearing configurations. Examples are included to demonstrate calculations.
This document discusses the kinematics of particles in rectilinear and curvilinear motion. It defines key concepts like position, displacement, velocity, and acceleration for both continuous and erratic rectilinear motion. Examples are provided to demonstrate how to construct velocity-time and acceleration-time graphs from a given position-time graph, and vice versa. The chapter then discusses general curvilinear motion, defining position, displacement, velocity, and acceleration using vector analysis since the curved path is three-dimensional. Fundamental problems and practice problems are also included.
This document provides information about the Strength of Materials CIE 102 course for first year B.E. degree students. It includes a list of 10 topics that will be covered in the course, such as simple stress and strain, shearing force and bending moment, and stability of columns. It also lists several reference books for the course and provides an overview of concepts that will be discussed in the first chapter, including stress, strain, stress-strain diagrams, and ductile vs brittle materials.
Any mechanical system can be classified by the number of degrees of freedom (DOF) it possesses, which is equal to the number of independent parameters needed to uniquely define its position in space at any given time. A DOF is defined relative to a selected frame of reference. The document discusses determining a mechanism's DOF using Kutzback's equation and provides examples of mechanisms with one and zero DOF that contain different types of joints.
The document discusses various topics related to stress and strain including: principal stresses and strains, Mohr's stress circle theory of failure, 3D stress and strain, equilibrium equations, and impact loading. It provides examples of stresses acting in different planes including normal, shear, oblique, and principal planes. It also gives examples of calculating normal and tangential stresses on an oblique plane subjected to stresses in one, two, or multiple directions with and without shear stresses.
The document discusses compound stresses, which involve both normal and shear stresses acting on a plane. It provides equations to calculate:
1) Normal and shear stresses on a plane inclined to the given stress plane.
2) The inclination and normal stresses on the planes of maximum and minimum normal stress (principal planes).
3) The inclination and shear stresses on the planes of maximum shear stress.
It includes an example problem calculating the principal stresses and maximum shear stresses given a state of stress. Sign conventions for stresses are also defined.
Overhanged Beam and Cantilever beam problemssushma chinta
This document discusses shear force diagrams (SFD) and bending moment diagrams (BMD) for overhanging beams and cantilever beams under different loading conditions. It provides examples of overhanging beams with uniform distributed loading and analyses the reactions, shear forces, bending moments, and point of contraflexure. It also discusses cantilever beams and provides examples of cantilevers with point loads and uniform distributed loads, deriving the corresponding SFDs and BMDs.
The document discusses various types of frictional clutches, including disc clutches, cone clutches, and centrifugal clutches. It provides details on the structure and operating principles of single plate and multi-plate disc clutches. The torque transmission analysis for these clutches is presented based on both uniform pressure and uniform wear theories. Cone clutches are described as providing higher torque transfer compared to disc clutches of the same size. Centrifugal clutches are explained as using centrifugal force to automatically engage and disengage based on engine speed.
Torsion (Springs)-Strength of Materials.pdfDr.S.SURESH
This document discusses different types of springs, including helical and leaf springs. Helical springs are made of coiled wire and store energy through torsion or twisting. Leaf springs are laminated and absorb shocks in vehicles. The document provides equations to calculate spring deflection, stiffness, and maximum shear/bending stress. It includes examples solving for the number of plates in a leaf spring, length of a leaf spring, and deflection under a given load.
1) Simple harmonic motion is the motion of an object where the acceleration is directly proportional to the displacement from the equilibrium position and directed towards the equilibrium.
2) It can be modeled as circular motion where the acceleration towards the center is proportional to the displacement from the center.
3) Simple harmonic oscillators include spring-mass systems and pendulums, where the restoring force is proportional to the displacement.
1) Simple harmonic motion (SHM) is a type of periodic motion where an object moves back and forth over the same path, like a mass on a spring or a pendulum.
2) For motion to be SHM, there must be a restoring force acting towards the equilibrium position that is proportional to the displacement.
3) The acceleration during SHM is directly proportional to the displacement from the equilibrium position and always acts to restore the object towards equilibrium.
Theory of machines_static and dynamic force analysisKiran Wakchaure
This document contains information about static and dynamic force analysis in machines. It discusses various topics including:
1) Types of forces such as static loads, dynamic loads, tension, compression, shear force, and torsion.
2) Laws of motion including Newton's three laws of motion.
3) Moment of inertia which is a mass property that determines the torque needed for angular acceleration. It depends on the shape and mass distribution of an object.
4) Analysis of simple and compound pendulums including calculations of their periodic times and frequencies of oscillation based on length, mass, and radius of gyration.
This document provides an overview of simple harmonic motion and pendulums. It includes:
- Objectives of understanding oscillations, deriving laws of oscillations, using Hooke's Law, and understanding simple harmonic motion.
- Experiments with a simple pendulum to determine factors affecting its period, including recording time for oscillations while varying length and mass.
- Theoretical analysis showing the period of a simple pendulum is proportional to the square root of its length.
- Deriving and solving the differential equation of motion for a simple pendulum using energy considerations.
- Examples of other oscillations that can be modeled using the derived simple harmonic motion equation.
This document provides information about simple harmonic motion over 6 hours. It begins by defining simple harmonic motion as periodic motion where the acceleration is directly proportional to and opposite of the displacement from equilibrium. Key equations for displacement, velocity, acceleration, kinetic energy and potential energy in SHM are given. Examples of simple harmonic motion include a simple pendulum and spring oscillations. The chapter then discusses the kinematics of SHM, deriving equations for displacement, velocity, acceleration and energies as functions of time and angular frequency.
1) Simple harmonic motion describes any oscillatory motion where the restoring force is directly proportional to the displacement. It follows the differential equation d2x/dt2 = -ω2x, where ω is the angular frequency.
2) The position as a function of time for simple harmonic motion is given by x(t) = Acos(ωt + φ), which describes a simple sinusoidal oscillation.
3) When a damping force is present that is proportional to the velocity, the motion is described by damped harmonic motion. The amplitude decreases exponentially with time in underdamped systems.
This document describes experiments performed with a torsion pendulum to study damped and driven oscillations. It summarizes the key characteristics of underdamped, critically damped, and overdamped oscillations. The experiments measured the damping constant and resonance frequency for different damping currents. While results for damping constants were unreliable due to insufficient data near resonance, the phase shift measurements matched expectations, increasing from 0 to 180 degrees with driving frequency and being 90 degrees at resonance. Improving data collection near resonance was suggested to obtain more accurate results.
The document discusses simple harmonic motion (SHM), where a particle moves back and forth such that its acceleration is directly proportional to its distance from a fixed point. SHM has the properties that the particle's velocity is zero at the amplitude of its oscillation, and it travels between the points of -a and +a on its axis. Examples are given of particles moving according to SHM equations, and exercises are provided to practice determining SHM, periods of motion, and maximum values.
Physics team at kibasila sec school made an animation on the Hooke's law of Elasticity. These teachers have never used a computer before for teaching purposes. This lesson is a result of a two days workshop and collaboration in design teams
This document discusses solving the harmonic oscillator equation to model different types of vibrations. It covers undamped free vibrations which exhibit simple harmonic motion. Damped free vibrations are also examined, where damping causes the amplitude to decay over time. Forced vibrations, including cases of beats and resonance, are explored. The document suggests the beam vibration can be modeled as a harmonic oscillator. It shows how to write the second order differential equation as a first order system for numerical solution in Matlab. Finally, it notes that the solution depends on ratios of m, c, and k, not their individual values, which is important for solving the inverse problem.
Simple harmonic motion describes back-and-forth motion caused by a restoring force proportional to displacement from equilibrium. Springs obey Hooke's law, where force is proportional to displacement. The period of a spring or pendulum undergoing simple harmonic motion can be determined from its total energy and amplitude or from equations relating displacement to velocity in circular motion.
A brief and easy concept of Simple harmonic oscillator. How we can get simple harmonic motion equation from Lagrange's equation of motion. How can we obtain this from Lagrange's equation of motion.
This document discusses simple harmonic motion and compares oscillating springs and pendulums. It defines simple harmonic motion as vibration about an equilibrium position where a restoring force is proportional to displacement. Springs and pendulums are given as examples that exhibit simple harmonic motion. The restoring force of a spring is described by Hooke's law as being proportional to displacement. For both springs and pendulums, the period of oscillation depends only on intrinsic properties and is independent of amplitude.
NIR spectroscopy is a technique that is widely used in pharmaceutical applications such as raw material identification, process monitoring, and finished product analysis. It works by measuring overtones and combinations of vibrational bonds like C-H, O-H, and N-H. Common instrumentation includes light sources, monochromators, sample holders, and detectors like PbS, PbSe, Si, InSb, and CCD. Applications include raw material and intermediate identification, tablet and capsule analysis, monitoring of processes like blending and coating, and agricultural uses like determining crop quality and chemical composition. Lyophilized products and final packaging can also be analyzed using NIR to ensure quality and identity.
This document outlines Tibor Astrab's investigation of simple harmonic motion. The experiment involves attaching different masses to a spring and measuring how the spring's oscillations change. Key points covered include:
1) The background physics of simple harmonic motion, including the exchange of potential and kinetic energy during oscillations and relevant equations.
2) The preliminary apparatus, method, and risk assessment, which will use similar equipment to the full experiment but with less precision.
3) Plans to measure the spring's stiffness constant and the effect of mass on the oscillation period.
4) Preliminary expectations for results graphs and evaluations of measurements.
5) An outline of the main apparatus, method, and risk assessment
Simple harmonic motion (SHM) refers to the periodic oscillatory motion of an object where the restoring force is directly proportional to the displacement of the object from its equilibrium position. There are two types of SHM: linear SHM such as oscillations of a spring or pendulum, and angular SHM such as a torsional pendulum. The period of SHM is the time required for one complete oscillation, while the frequency is the number of oscillations that occur per second. For a spring with spring constant k and mass m, the period is 2π√(m/k). For a simple pendulum of length l, the period is 2π√(l/g). SHM can be described by equations
Pendulum Lap investigating the relationship between the length of the pendlum string and the time needed for the oscillations
Score archieved: 5/6 in the DCP section.
This document discusses simple harmonic motion (SHM) and related concepts like angular velocity, restoring forces, displacement, velocity, acceleration, energy, resonance, and damping. It provides equations for angular velocity, SHM acceleration, displacement, velocity, energy, and the simple pendulum. Examples are given and questions provided for practice calculations involving these equations and concepts.
This document summarizes key concepts about simple harmonic motion:
1) Simple harmonic motion occurs when the acceleration of an object is proportional to and opposite of the displacement from a central point. The force causing the motion is called a restoring force.
2) Periodic variables that describe simple harmonic motion include amplitude, period, frequency, and angular frequency. The period is the time for one complete oscillation.
3) Examples of simple harmonic motion include a mass on a spring, the motion of a simple pendulum, and uniform circular motion. For a mass on a spring, the acceleration is proportional to the displacement from equilibrium.
4) Other types of oscillatory motion discussed include the simple pendulum and
This document provides an overview of natural vibrations with one degree of freedom. It defines key terms like degrees of freedom, simple harmonic motion, angular frequency, and periodic time. Examples of natural vibrations covered include a simple pendulum, a mass on a spring, and beams experiencing transverse vibrations. Methods for analyzing natural vibrations include Rayleigh's energy method and Dunkerley's method for combining frequencies from distributed and point loads. Worked examples are provided to illustrate calculating displacement, velocity, and acceleration of bodies in simple harmonic motion.
1) The document discusses kinematics of particles, specifically rectilinear and curvilinear motion.
2) Rectilinear motion involves motion along a straight line that can be described using displacement, velocity, acceleration, and differential equations.
3) Curvilinear motion occurs along a curved path in a plane and uses vector analysis to describe position, displacement, and velocity.
1. Circular motion involves an object moving in a circular path at a constant speed. Centripetal force is required to cause an object to travel in a circular path and undergo centripetal acceleration towards the center.
2. Key concepts of circular motion include: angular velocity, linear velocity, centripetal acceleration, centripetal force, period and frequency. Formulas relate these concepts, such as v=rw, a=v^2/r.
3. Examples of circular motion include planets orbiting the sun, electrons orbiting an atom's nucleus, objects on the end of a string being swung in a circle. Applications involve calculating speeds, tensions, and accelerations in situations like a t
This document discusses rotational motion and provides definitions and equations for key angular quantities such as angular displacement (θ), angular velocity (ω), angular acceleration (α), torque (τ), moment of inertia (I), angular momentum (L), and rotational kinetic energy. It defines these quantities, gives their relationships to linear motion quantities, and provides examples of how to set up and solve problems involving rotational dynamics.
1. Simple harmonic motion is motion influenced by a restoring force proportional to displacement from equilibrium. For a spring, F=-kx, and for a pendulum, F=mg sinθ.
2. The period T of a spring is the time for one complete oscillation, related to spring constant k and mass m by T=2π√(m/k). Frequency f is the number of oscillations per second, with f=1/T.
3. The displacement y of a spring over time t follows a sinusoidal pattern described by the equation y=A sin(2πft+θ0), where A is the amplitude
PHYSICS (CLASSS XII) - Chapter 5 : OscillationsPooja M
1. Oscillatory motion is a periodic motion that repeats itself after a definite time interval called the period. Linear simple harmonic motion (S.H.M.) is the linear periodic motion of a body where the restoring force is directly proportional to the displacement from the mean position.
2. The differential equation of S.H.M is the second derivative of displacement (x) plus the angular frequency (ω) squared times x equals zero. The expressions for displacement, velocity, and acceleration of a particle in S.H.M involve sinusoidal and cosine functions of angular frequency times time.
3. The amplitude is the maximum displacement, the period is the time for one oscillation, and the frequency is
Wk 1 p7 wk 3-p8_13.1-13.3 & 14.6_oscillations & ultrasoundchris lembalemba
This document discusses oscillations and simple harmonic motion. It begins by listing learning outcomes related to oscillations, including describing examples of free and forced oscillations. It then provides definitions and equations related to simple harmonic motion, such as the defining equation a=-ω2x. Graphs of displacement, velocity, and acceleration over time for simple harmonic motion are shown. Examples of the simple pendulum and a mass on a spring are provided to illustrate simple harmonic motion. The document also discusses the kinetic and potential energy changes that occur during simple harmonic motion.
13.1.1 Shm Part 1 Introducing Circular MotionChris Staines
1. Bodies moving in circular motion are constantly accelerating towards the center of the circle due to the centripetal force.
2. The centripetal acceleration of an object can be calculated as a = ω2r, where ω is the angular velocity and r is the radius of the circular path.
3. Deriving this relationship involves considering the change in velocity between two nearby points on the circular path and relating this to the arc length subtended and the time interval between the two points.
1. The lecture goals were to describe oscillations and simple harmonic motion, analyze them using energy concepts, and apply SHM to different physical situations like pendulums and driving forces.
2. The document then covered topics like equilibrium, restoring forces, characteristics of periodic motion, and the mathematics of simple harmonic oscillators.
3. It concluded by discussing mechanical waves, including transverse and longitudinal waves, wave speed, interference, and standing waves on a string.
This document discusses simple harmonic motion (SHM). SHM occurs when an object experiences a restoring force proportional to its displacement from equilibrium. This results in sinusoidal oscillations described by x(t) = Acos(ωt + φ), where A is amplitude, ω is angular frequency, and φ is the phase. SHM includes examples like a mass on a spring and a simple pendulum. The relationships between displacement, velocity, acceleration, period, frequency, and energy in SHM systems are explored.
GEN PHYSICS 1 WEEK 2 KINEMATICS IN ONE DIMENSION.pptxAshmontefalco4
This document provides an overview of kinematics in one dimension, including graphical representations of motion, motion along a straight line, and motion with constant acceleration. It discusses topics like displacement, velocity, acceleration, and their relationships. Examples are provided to demonstrate calculating variables like position, velocity and acceleration from graphs or equations of motion. Kinematic formulas are also introduced for problems involving constant acceleration.
Introduction to oscillations and simple harmonic motionMichael Marty
Physics presentation about Simple Harmonic Motion of Hooke's Law springs and pendulums with derivation of formulas and connections to Uniform Circular Motion.
References include links to illustrative youtube clips and other powerpoints that contributed to this peresentation.
This document defines key terms and equations related to simple harmonic motion (SHM). It discusses oscillating systems that vibrate back and forth around an equilibrium point, like a mass on a spring or pendulum. The key parameters of SHM systems are defined, including amplitude, wavelength, period, frequency, displacement, velocity, acceleration. Equations are presented that relate the displacement, velocity, acceleration as sinusoidal functions of time. The concepts of kinetic, potential and total energy are also explained for oscillating systems undergoing SHM.
The document discusses kinematics of particles, including rectilinear and curvilinear motion. It defines key concepts like displacement, velocity, and acceleration. It presents equations for calculating these values for rectilinear motion under different conditions of acceleration, such as constant acceleration, acceleration as a function of time, velocity, or displacement. Graphical interpretations are also described. An example problem is worked through to demonstrate finding velocity, acceleration, and displacement at different times for a particle moving in a straight line.
1. The document discusses uniform circular motion, defining key terms like linear velocity, angular velocity, centripetal force, and centripetal acceleration.
2. Examples of circular motion are given, and the relationships between linear velocity, angular velocity, radius, period, and frequency are defined.
3. Centripetal force is described as the force directing an object towards the center of its circular path, and equations are provided relating centripetal force to mass, velocity, and radius.
International Journal of Computational Engineering Research(IJCER)ijceronline
The document discusses using an elliptical curve as an alternative to compound or spiral curves for highway horizontal alignments. It provides equations to define an ellipse based on its major and minor axes and eccentricity. An approach is presented to determine the elliptical curve that satisfies design speed and radius requirements for a given highway design problem. Algorithms are provided to calculate chord lengths, deflection angles, and station points along the elliptical curve. An example problem applies the elliptical curve approach and compares results to equivalent circular and spiral-circular curves.
Here are the key steps to solve this problem:
1) Find the linear acceleration (a = 0.800 m/s2)
2) Find the time of acceleration (t = 20.0 s)
3) Use the equation for linear acceleration (a = rα) to find the angular acceleration:
a / r = α
0.800 m/s2 / 0.330 m = α
α = 2.42 rad/s2
4) Use the equation for angular velocity (ω = ω0 + αt) to find the final angular velocity:
ω = 0 + 2.42 rad/s2 * 20.0 s
ω = 48.4 rad/s
1. 72 l Theory of Machines
Simple
Harmonic
Motion
4Features
1. Introduction.
2. Velocity and Acceleration of
a Particle Moving with
Simple Harmonic Motion.
3. Differential Equation of
Simple Harmonic Motion.
4. Terms Used in Simple
Harmonic Motion.
5. Simple Pendulum.
6. Laws of Simple Pendulum.
7. Closely-coiled Helical
Spring.
8. Compound Pendulum.
9. Centre of Percussion.
10. Bifilar Suspension.
11. Trifilar Suspension
(Torsional Pendulum).
4.1. Introduction
Consider a particle
moving round the circumfer-
ence of a circle in an
anticlockwise direction, with
a constant angular velocity,
as shown in Fig. 4.1. Let P
be the position of the particle
at any instant and N be the
projection of P on the diam-
eter X X ′ of the circle.
It will be noticed that when
the point P moves round the
circumference of the circle from X to
Y, N moves from X to O, when P
moves from Y to X ′, N moves from O
to X ′. Similarly when P moves from
X ′ to Y ′, N moves from X ′ to O and
finally when Pmoves from Y ′ to X, N
moves from O to X. Hence, as P
completes one revolution, the point N
completes one vibration about the
Fig. 4.1. Simple harmonic
motion.
A clock pendulum
executes Simple
Harmonic Motion.
72
CONTENTSCONTENTS
CONTENTSCONTENTS
2. Chapter 4 : Simple Harmonic Motion l 73
point O. This to and fro motion of N is known as simple
harmonic motion (briefly written as S.H.M.).
4.2. Velocity and Acceleration of a
Particle Moving with Simple
Harmonic Motion
Consider a particle, moving round the circumfer-
ence of a circle of radius r, with a uniform angular velocity
ω rad/s, as shown in Fig. 4.2. Let P be any position of the
particle after t seconds and θ be the angle turned by the
particle in t seconds. We know that
θ = ω.t
If N is the projection of P on the diameter X X ′,
then displacement of N from its mean position O is
x = r.cos θ = r.cos ω.t ... (i)
The velocity of N is the component of the velocity
of P parallel to XX ′, i.e.
2 2
N sin . sinv v r r x= θ = ω θ = ω − ... (ii)
2 2
... . , and sinv r r NP r x = ω θ = = −
3
Fig. 4.2. Velocity and acceleration of a particle.
A little consideration will show that velocity is maximum, when x = 0, i.e. when N passes
through O i.e., its mean position.
∴ vmax = ω.r
We also know that the acceleration of P is the centripetal acceleration whose magnitude is
ω2.r. The acceleration of N is the component of the acceleration of P parallel to XX ′ and is directed
towards the centre O, i.e.,
2 2
N . cos .a r x= ω θ = ω ... ( cos )x r= θ3 ...(iii)
The acceleration is maximum when x = r i.e. when P is at X or X′.
∴ amax = ω2.r
It will also be noticed from equation (iii) that when x = 0, the acceleration is zero i.e. N passes
through O. In other words, the acceleration is zero at the mean position. Thus we see from equation
(iii) that the acceleration of N is proportional to its displacement from its mean position O, and it is
Movements of a ship up and down in
a vertical plane about transverse axis
(called Pitching) and about longitude
(called rolling) are in Simple
Harmonic Motion.
3. 74 l Theory of Machines
*
always directed towards the centre O; so that the motion of N is simple harmonic.
In general, a body is said to move or vibrate with simple harmonic motion, if it satisfies the
following two conditions :
1. Its acceleration is always directed towards the centre, known as point of reference or
mean position ;
2. Its acceleration is proportional to the distance from that point.
4.3. Differential Equation of Simple Harmonic Motion
We have discussed in the previous article that the displacement of N from its mean position O is
x = r.cos θ = r.cos ωt ... (i)
Differentiating equation (i), we have velocity of N,
N . sindx v r t
dt
= = ω ω ... (ii)
Again differentiating equation (ii), we have acceleration of N,
2
2 2
N2
. . cos . cos .
d x
a r t r t x
dt
= =− ω ω ω = − ω ω = − ω ... (iii)
... (3 r cos ωt = x)
or
2
2
2
0d x x
dt
+ ω =
This is the standard differential equation for simple harmonic motion of a particle. The
solution of this differential equation is
x = A cos ω t + B sin ω t ... (iv)
where A and B are constants to be determined by the initial conditions of the motion.
In Fig. 4.2, when t = 0, x = ri.e. when points P and N lie at X, we have from equation (iv), A = r
Differentiating equation (iv),
. .sin . cosdx A t B t
dt
= − ω ω + ω ω
When 0, 0,dxt
dt
= = therefore, from the above equation, B = 0. Now the equation (iv) becomes
x = r cos ω t . . . [Same as equation (i)]
The equations (ii) and (iii) may be written as
N . sin . cos ( / 2)dx v r t r t
dt
= = − ω ω = ω ω + π
and
2
2 2
N2
. cos . cos( )d x a r t r t
dt
= = − ω ω = ω ω + π
These equations show that the velocity leads the displacement by 90° and acceleration leads
the displacement by 180°.
* The negative sign shows that the direction of acceleration is opposite to the direction in which x increases,
i.e. the acceleration is always directed towards the point O.
4. Chapter 4 : Simple Harmonic Motion l 75
4.4. Terms Used in Simple Harmonic Motion
The following terms, commonly used in simple harmonic motion, are important from the
subject point of view.
1. Amplitude. It is the maximum displacement of a body from its mean position. In Fig. 4.2,
OX or OX ′ is the amplitude of the particle P. The amplitude is always equal to the radius of the
circle.
2. Periodic time. It is the time taken for one complete revolution of the particle.
∴ Periodic time, tp = 2 π/ω seconds
We know that the acceleration,
a = ω2.x or 2
ora a
x x
ω = ω =
∴ p
Displacement2 2 2 seconds
Acceleration
xt
a
π= = π = π
ω
It is thus obvious, that the periodic time is independent of amplitude.
3. Frequency. It is the number of cycles per second and is the reciprocal of time period, tp .
∴ Frequency, 1 1 Hz
2 2p
an
t x
ω
= = =
π π
Notes : 1. In S.I. units, the unit of frequency is hertz (briefly written as Hz) which is equal to one cycle per
second.
2. When the particle moves with angular simple harmonic motion, then the periodic time,
p
Angular displacement
2 2 s
Angular acceleration
t θ= π = π
α
and frequency,
1
Hz
2
n
α
=
π θ
Example 4.1. The piston of a steam engine moves with simple harmonic motion. The crank
rotates at 120 r.p.m. with a stroke of 2 metres. Find the velocity and acceleration of the piston, when
it is at a distance of 0.75 metre from the centre.
Solution. Given : N = 120 r.p.m. or ω = 2π × 120/60 = 4π rad/s ; 2r = 2 m or r = 1 m;
x = 0.75 m
Velocity of the piston
We know that velocity of the piston,
2 2 2
4 1 (0.75) 8.31 m/sv r x= ω − = π − = Ans.
Acceleration of the piston
We also know that acceleration of the piston,
a = ω2.x = (4π)2 0.75 = 118.46 m/s2 Ans.
Example 4.2. A point moves with simple harmonic motion. When this point is 0.75 metre
from the mid path, its velocity is 11 m/s and when 2 metres from the centre of its path its velocity is
3 m/s. Find its angular velocity, periodic time and its maximum acceleration.
5. 76 l Theory of Machines
Solution. Given : When x = 0.75 m, v = 11 m/s ; when x = 2 m, v = 3 m/s
Angular velocity
Let ω = Angular velocity of the particle, and
r = Amplitude of the particle.
We know that velocity of the point when it is 0.75 m from the mid path (v),
2 2 2 2
11 (0.75)r x r= ω − = ω − . . . (i)
Similarly, velocity of the point when it is 2 m from the centre (v),
2 2
3 2r= ω − . . . (ii)
Dividing equation (i) by equation (ii),
2 2 2 2
2 2 2 2
(0.75) (0.75)11
3 2 2
r r
r r
ω − −
= =
ω − −
Squaring both sides,
2
2
0.5625121
9 4
r
r
−
=
−
121 r2 – 484 = 9r2 – 5.06 or 112 r2 = 478.94
∴ r2 = 478.94 / 112 = 4.276 or r = 2.07 m
Substituting the value of r in equation (i),
2 2
11 (2.07) (0.75) 1.93= ω − = ω
∴ ω = 11/1.93 = 5.7 rad/s Ans.
Periodic time
We know that periodic time,
tp = 2π / ω = 2π / 5.7 = 1.1 s Ans.
Maximum acceleration
We know that maximum acceleration,
amax = ω2.r = (5.7)2 2.07 = 67.25 m/s2 Ans.
4.5. Simple Pendulum
A simple pendulum, in its simplest form, consists
of heavy bob suspended at the end of a light inextensible
and flexible string. The other end of the string is fixed at
O, as shown in Fig. 4.3.
Let L = Length of the string,
m = Mass of the bob in kg,
W = Weight of the bob in newtons
= m.g, and
θ = Angle through which the string
is displaced. Fig 4.3. Simple pendulum.
6. Chapter 4 : Simple Harmonic Motion l 77
When the bob is at A, the pendulum is in equilibrium position. If the bob is brought to B or C
and released, it will start oscillating between the two positions B and C, with A as the mean position.
It has been observed that if the angle θ is very small (less than 4° ), the bob will have simple harmonic
motion. Now, the couple tending to restore the bob to the equilibrium position or restoring torque,
T = m.g sin θ × L
Since angle θ is very small, therefore sin θ = θ radians.
∴ T = m.g.L.θ
We know that the mass moment of inertia of the bob about an axis through the point of
suspension,
I = mass × (length)2 = m.L2
∴ Angular acceleration of the string,
2
. . . .
or
.
m g L gT L
I L gm L
θ θ θα = = = =
α
i.e.
Angular displacement
Angular acceleration
L
g
=
We know that the periodic time,
Displacement
2 2
Accelerationp
Lt
g
= π = π ... (i)
and frequency of oscillation,
1 1
2p
g
n
t L
= =
π ... (ii)
From above we see that the periodic time and the frequency of oscillation of a simple
pendulum depends only upon its length and acceleration due to gravity. The mass of the bob has no
effect on it.
Notes : 1. The motion of the bob from one extremity to the other (i.e. from B to C or C to B) is known as beat
or swing. Thus one beat = 1
2
oscillation.
∴ Periodic time for one beat = L gπ /
2. A pendulum, which executes one beat per second (i.e. one complete oscillation in two seconds) is
known as a second’s pendulum.
4.6. Laws of Simple Pendulum
The following laws of a simple pendulum are important from the subject point of view :
1. Law of isochronism. It states, “The time period (tp ) of a simple pendulum does not depend
upon its amplitude of vibration and remains the same, provided the angular amplitude (θ) does not
exceed 4°.”
2. Law of mass. It states, “The time period (tp ) of a simple pendulum does not depend upon
the mass of the body suspended at the free end of the string.”
3. Law of length. It states, “The time period (tp ) of a simple pendulum is directly propor-
tional to L , where L is the length of the string.”
4. Law of gravity. It states, “The time period (tp ) of a simple pendulum is inversely propor-
tional to g , where g is the acceleration due to gravity.”
7. 78 l Theory of Machines
Fig. 4.4. Closely-coiled
helical spring.
* The differential equation for the motion of the spring is
2 2
2 2
.. or –
d x d x s x
m s x
mdt dt
= − = ... ( )2
Here
s
m
ω =
The – ve sign indicates that the restoring force s.x is opposite to the direction of disturbing force.
Note: The above laws of a simple pendulum are true from the equation of the periodic time i.e.
2 /pt L g= π
4.7. Closely-coiled Helical Spring
Consider a closely-coiled helical spring, whose upper end is
fixed, as shown in Fig. 4.4. Let a body be attached to the lower end.
Let A A be the equilibrium position of the spring, after the mass is
attached. If the spring is stretched up to BB and then released, the
mass will move up and down with simple harmonic motion.
Let m = Mass of the body in kg,
W = Weight of the body in newtons = m.g,
x = Displacement of the load below equilib-
rium position in metres,
s = Stiffnes of the spring in N/m i.e. restoring
force per unit displacement from the equi-
librium position,
a = Acceleration of the body in m/s2.
We know that the deflection of the spring,
.m g
s
δ = ... (i)
Then disturbing force = m.a
and restoring force = s.x ... (ii)
Equating equations (i) and (ii),
m.a = s.x* or =x m
a s
8. Chapter 4 : Simple Harmonic Motion l 79
We know that periodic time,
Displacement
2 2
Accelerationp
xt
a
= π = π
2 2
m
s g
δ= π = π ...
mg
s
δ =
3
and frequency, 1 1 1
2 2p
gsn
t m
= = =
π π δ
Note: If the mass of the spring (m1) is also taken into consideration, then the periodic time,
1 /3
2p
m m
t
s
+
= π seconds,
and frequency,
1
1 Hz
2 / 3
sn
m m
=
π +
Example 4.3. A helical spring, of negligible mass, and which is found to extend 0.25 mm
under a mass of 1.5 kg, is made to support a mass of 60 kg. The spring and the mass system is
displaced vertically through 12.5 mm and released. Determine the frequency of natural vibration of
the system. Find also the velocity of the mass, when it is 5 mm below its rest position.
Solution. Given : m = 60 kg ; r = 12.5 mm = 0.0125 m ; x = 5 mm = 0.005 m
Since a mass of 1.5 kg extends the spring by 0.25 mm, therefore a mass of 60 kg will extend
the spring by an amount,
0.25 60 10 mm = 0.01 m
1.5
δ = × =
Frequency of the system
We know that frequency of the system,
1 1 9.81
4.98
2 2 0.01
g
n = = =
π δ π
Hz Ans.
Velocity of the mass
Let v = Linear velocity of the mass.
We know that angular velocity,
ω*
9.81
* 31.32
0.01
g
= = =
δ
rad/s
and
2 2 2 2
31.32 (0.0125) (0.005)v r x=ω − = − = 0.36 m/s Ans.
4.8. Compound Pendulum
When a rigid body is suspended vertically, and it oscillates
with a small amplitude under the action of the force of gravity, the
body is known as compound pendulum, as shown in Fig. 4.5.
Let m = Mass of the pendulum in kg,
W = Weight of the pendulum in
newtons = m.g,
Fig. 4.5. Compound pendulum.
* We know that periodic time,
tp = 2π / ω or ω = 2π / tp = 2π × n = 2π × 4.98 = 31.3 rad/s ...(∵ n = 1/tp)
9. 80 l Theory of Machines
kG = Radius of gyration
about an axis
through the centre
of gravity G and
perpendicular to
the plane of
motion, and
h = Distance of point
of suspension O
from the centre of
gravity G of the
body.
If the pendulum is given a small
angular displacement θ, then the couple
tending to restore the pendulum to the
equilibrium position OA,
T = mg sin θ × h = mgh sin θ
Since θ is very small, therefore sub-
stituting sin θ = θ radians, we get
T = mgh θ
Now, the mass moment of inertia about the axis of suspension O,
( )2 2 2
G G.I I m h m k h= + = + . . . (By parallel axis theorem)
∴ Angular acceleration of the pendulum,
2 2 2 2
G G( )
mgh ghT
I m k h k h
θ θ
α = = =
+ +
= constant × θ
We see that the angular acceleration is directly proportional to angular displacement,
therefore the pendulum executes simple harmonic motion.
∴
2 2
G
.
k h
g h
+θ =
α
We know that the periodic time,
Displacement
2 2
Accelerationpt
θ= π = π
α
2 2
G
2
.
k h
g h
+
= π ... (i)
and frequency of oscillation, 2 2
G
.1 1
2p
g h
n
t k h
= =
π +
... (ii)
10. Chapter 4 : Simple Harmonic Motion l 81
Notes : 1. Comparing this equation with equation (ii) of simple pendulum, we see that the equivalent length of
a simple pendulum, which gives the same frequency as compound pendulum, is
2 2 2
G Gk h k
L h
h h
+
= = +
2. Since the equivalent length of simple pendulum (L) depends upon the distance between the point of
suspension and the centre of gravity (G), therefore L can be changed by changing the position of point of suspen-
sion. This will, obviously, change the periodic time of a compound pendulum. The periodic time will be minimum
if L is minimum. For L to be minimum, the differentiation of L with respect to h must be equal to zero, i.e.
2
G
0 or 0
kdL d h
dh dh h
= + =
∴
2
G
2
1 0
k
h
−
+ = or kG = h
Thus the periodic time of a compound pendulum is minimum when the distance between the point of
suspension and the centre of gravity is equal to the radius of gyration of the body about its centre of gravity.
∴ Minimum periodic time of a compound pendulum,
G
( )
2
2p min
k
t
g
= π . . . [Substituting h = kG in equation (i)]
4.9. Centre of Percussion
The centre of oscillation is sometimes termed as cen-
tre of percussion. It is defined as that point at which a blow
may be struck on a suspended body so that the reaction at the
support is zero.
Consider the case of a compound pendulum suspended
at O as shown in Fig. 4.6. Suppose the pendulum is at rest in
the vertical position, and a blow is struck at a distance L from
the centre of suspension. Let the magnitude of blow is F new-
tons. A little consideration will show that this blow will have
the following two effects on the body :
1. A force (F) acting at C will produce a linear motion
with an acceleration a, such that
F = m.a ... (i)
where m is the mass of the body.
2. A couple with moment equal to (F × l ) which will tend to produce a motion of rotation in
the clockwise direction about the centre of gravity G. Let this turning moment (F × l) produce an
angular acceleration (α), such that
F × l = IG × α ... (ii)
where IG is the moment of inertia of the body about an axis passing through G and parallel to the axis
of rotation.
From equation (i) a = F/m ... (iii)
and from equation (ii),
G
.F l
I
α =
Fig. 4.6. Centre of percusssion.
11. 82 l Theory of Machines
Now corresponding linear acceleration of O,
0 2
G G
. . . .
.
F l h F l ha h
I m k
= α. = = ... (iv)
2
G G( . )I m k=3
where kG is the radius of gyration of the body about the centre of
gravity G.
Since there is no reaction at the support when the body is
struck at the centre of percussion, therefore a should be equal to a0.
Equating equations (iii) and (iv),
2
G
. .
.
F F l h
m m k
=
or
2
2 G
G . , and
k
k l h l
h
= = ... (v)
We know that the equivalent length of a simple pendulum,
2 2
G Gk h k
L h l h
h h
+
= = + = + ... (vi)
From equations (v) and (vi), it follows that
1. The centre of percussion is below the centre of gravity
and at a distance 2
G / .k h
2. The distance between the centre of suspension and the centre of percussion is equal to the
equivalent length of a simple pendulum.
Note: We know that mass moment of inertia of the body about O,
2 2 2 2
O G G O. or . . .I I m h m k m k m h= + = +
∴ 2 2 2 2
O G . ( )k k h l h h h l h OG OC= + = + = + = × ... 2
G( . )k l h=3
It is thus obvious that the centre of suspension (O) and the centre of percussion (C) are inter-changeable.
In other words, the periodic time and frequency of oscillation will be same, whether the body is suspended at the
point of suspension or at the centre of percussion.
Example 4.4. A uniform thin rod, as shown in Fig. 4.7, has a mass of 1 kg and carries a
concentrated mass of 2.5 kg at B. The rod is hinged at A and is maintained in the horizontal position
by a spring of stiffness 1.8 kN/m at C.
Find the frequency of oscillation, neglecting the effect of the mass of the spring.
Fig. 4.7
A pendulum clock designed
by Galileo. Galileo was the
first to deisgn a clock based
on the relationship between
gravitational force (g), length
of the pendulum (l ) and time
of oscillation (t).
12. Chapter 4 : Simple Harmonic Motion l 83
Solution. Given : m = 1 kg ; m1 = 2.5 kg ; s = 1.8 kN/m = 1.8 × 103 N/m
We know that total length of rod,
l = 300 + 300 = 600 mm = 0.6 m
∴ Mass moment of inertia of the system about A,
IA = Mass moment of inertia of 1 kg about A + Mass moment of interia of
2.5 kg about A
22
2 2
1
1(0.6). . 2.5 (0.6) 1.02
3 3
ml m l= + = + = kg-m2
If the rod is given a small angular displacement θ and then released, the extension of the spring,
δ = 0.3 sin θ = 0.3θ m
. . . ( ∵ θ is very small, therefore substituting sin θ = θ )
∴ Restoring force = s.δ = 1.8 × 103 × 0.3 θ = 540 θ N
and restoring torque about A = 540 θ × 0.3 = 162 θ N-m ... (i)
We know that disturbing torque about A
= IA × α = 1.02α N-m ... (ii)
Equating equations (i) and (ii),
1.02 α = 162 θ or α / θ = 162 / 1.02 = 159
We know that frequency of oscillation,
1 1 159
2 2
n α= =
π θ π
= 2.01 Hz Ans.
Example 4.5. A small flywheel of mass 85 kg is suspended in a vertical plane as a compound
pendulum. The distance of centre of gravity from the knife edge support is 100 mm and the flywheel
makes 100 oscillations in 145 seconds. Find the moment of inertia of the flywheel through the centre
of gravity.
Solution. Given : m = 85 kg ; h = 100 mm = 0.1 m
Since the flywheel makes 100 oscillations in 145 seconds, therefore frequency of oscillation,
n = 100/145 = 0.69 Hz
Let L = Equivalent length of simple pendulum, and
kG = Radius of gyration through C.G.
We know that frequency of oscillation (n),
1 1 9.81 0.50.69
2 2
g
L L L
= = =
π π
∴ L = 0.5/0.69 = 0.7246 or L = 0.525 m
We also know that equivalent length of simple pendulum (L),
2 2 2 2
G G G (0.1)
0.525 0.1
0.1 0.1
k k k
h
h
+
= + = + =
2 2 2
G 0.525 0.1 (0.1) 0.0425 mk = × − =
13. 84 l Theory of Machines
and moment of inertia of the flywheel through the centre of gravity,
I =
2
G.m k = 85 × 0.0425 = 3.6 kg-m2 Ans.
Example 4.6. The connecting rod of an oil engine has a mass of 60 kg, the distance between
the bearing centres is 1 metre. The diameter of the big end bearing is 120 mm and of the small end
bearing is 75 mm. When suspended vertically with a knife-edge through the small end, it makes 100
oscillations in 190 seconds and with knife-edge through the big end it makes 100 oscillations in 165
seconds. Find the moment of inertia of the rod in kg-m2 and the distance of C.G. from the small end
centre.
Solution. Given : m = 60 kg ; h1 + h2 = 1 m ; d2* = 102 mm; d1* = 75 mm
Moment of inertia of the rod
First of all, let us find the radius of gyration of the connecting
rod about the centre of gravity (i.e. kG).
Let h1 and h2 = Distance of centre of gravity from
the small and big end centres respec-
tively,
L1 and L2 = Equivalent length of simple
perdulum when the axis of oscilla-
tion coincides with the small and big
end centres respectively
When the axis of oscillation coincides with the small end cen-
tre, then frequency of oscillation,
n1 = 100/190 = 0.526 Hz
When the axis of oscillation coincides with the big end centre, the frequency of oscillation,
n2 = 100/165 = 0.606 Hz
We know that for a simple pendulum,
1
1
1
Hz
2
g
n
L
=
π
∴ 1 2 2
1
9.81
0.9m
(2 ) (2 0.526)
g
L
n
= = =
π π ×
Similarly 2 2 2
2
9.81 0.67m
(2 ) (2 0.606)
g
L
n
= = =
π π ×
We know that
2 2
2 2G 1
1 G 1 1 1
1
( )
or . ( )
k h
L k L h h
h
+
= = − ... (i)
Similarly 2 2
G 2 2 2. ( )k L h h= − ... (ii)
From equations (i) and (ii), we have
L1.h1 – (h1)2 = L2.h2 – (h2)2
* Superfluous data.
Connecting rod
14. Chapter 4 : Simple Harmonic Motion l 85
0.9 × h1 – (h1)2 = 0.67 (1 – h1) – (1 – h1)2 . . . (3 h1 + h2 = 1 m)
= 0.67 – 0.67 h1 – 1 – (h1)2 + 2h1
0.9 h1 + 0.67 h1 – 2 h1 = – 0.33 or – 0.43 h1 = – 0.33
∴ h1 = 0.33/0.43 = 0.767 m
Substituting the value of h1 in equation (i), we have
2 2 2
G 0.9 0.767 (0.767) 0.69 0.59 0.1mk = × − = − =
We know that mass moment of inertia of the rod,
2
G.I m k= = 60 × 0.1 = 6 kg-m2 Ans.
Distance of C.G. from the small end centre
We have calculated above that the distance of C.G. from the small end centre,
h1 = 0.767 m Ans.
Example 4.7. A uniform slender rod 1.2 m long is fitted with a transverse pair of knife-
edges, so that it can swing in a vertical plane as a compound pendulum. The position of the knife
edges is variable. Find the time of swing of the rod, if 1. the knife edges are 50 mm from one end of
the rod, and 2. the knife edges are so placed that the time of swing is minimum.
In case (1) find also the maximum angular velocity and the maximum angular acceleration
of the rod if it swings through 3° on either side of the vertical.
Solution. Given : l = 1.2 m ; θ = 3° = 3 × π /180 = 0.052 rad
1. Time of swing of the rod when knife edges are 50 mm
Since the distance between knife edges from one end of the rod is 50 mm = 0.05 m, therefore
distance between the knife edge and C.G. of the rod,
1.2 0.05 0.55m
2
h = − =
We know that radius of gyration of the rod about C.G.,
k*
G
1.2
0.35 m
12 12
l
= = =
∴ Time of swing of the rod,
2 2 2 2
G (0.35) (0.55)
2 2
. 9.81 0.55p
k h
t
g h
+ +
= π = π
×
= 1.76 s Ans.
2. Minimum time of swing
We know that minimum time of swing,
G
( )
2 2 0.35
2 2 1.68 s
9.81p min
k
t
g
×
= π = π = Ans.
* We know that mass moment of inertia of the rod about an axis through C.G.
I = m. l2/12
Also I = m.k2 or k2 = I/m = m.l2/12 × m = l 2/12 or k = 12l
15. 86 l Theory of Machines
Maximum angular velocity
In case (1), the angular velocity,
ω = 2π / tp = 2π /1.76 = 3.57 rad/s
We know that maximum angular velocity,
ωmax = ω.θ = 3.57 × 0.052 = 0.1856 rad/s Ans.
Maximum angular acceleration
We know that maximum angular acceleration,
αmax = ω2.θ = (3.57)2 × 0.052 = 0.663 rad/s2 Ans.
Example 4.8. The pendulum of an Izod impact testing machine has a mass of 30 kg. Its
centre of gravity is 1.05 m from the axis of suspension and the striking knife is 150 mm below the
centre of gravity. The time for 20 small free oscillations is 43.5 seconds. In making a test the pendu-
lum is released from an angle of 60° to the vertical. Determine :
1. the position of the centre of percussion relative to the striking knife and the striking veloc-
ity of the pendulum, and 2. the impulse on the pendulum and the sudden change of axis reaction
when a specimen giving an impact value of 55 N-m is broken.
Solution. Given : m = 30 kg ; OG = h = 1.05 m ; AG = 0.15 m
Since the time for 20 small free oscillations is 43.5 s, therefore frequency of oscillation,
20 0.46 Hz
43.5
n = =
1. The position of centre of percussion relative to the striking knife and the striking velocity of the
pendulum
Let L = Equivalent length of simple pendulum,
kG = Radius of gyration of the pendulum about the centre of gravity, and
kO = Radius of gyration of the pendulum about O.
We know that the frequency of oscillation,
1
2
g
n
L
=
π
or 2 2
9.81
(2 ) (2 0.46)
g
L
n
= =
π π ×
= 1.174 m
∴ Distance of centre of percussion (C) from
the centre of gravity (G),
CG = OC – OG = L – OG
= 1.174 – 1.05 = 0.124 m
and distance of centre of percussion (C) from knife edge A,
AC = AG – CG = 0.15 – 0.124 = 0.026 m Ans.
We know that 2 2
O ( ) . 1.174 1.05 1.233 mk h l h L h= + = = × =
A little consideration will show that the potential energy of the pendulum is converted into
kinetic energy of the pendulum before it strikes the test piece. Let v and ω be the linear and angular
velocity of the pendulum before it strikes the test piece.
Fig. 4.8
16. Chapter 4 : Simple Harmonic Motion l 87
∴ m.g.h1 =
2 2 2
O
1 1. . .
2 2
m v m k= ω ... (∵ v = kO.ω)
30 × 9.81 × 1.05 (1 – cos 60°) =
21 30 1.233
2
× × ω or 154.5 = 18.5 ω2
∴ ω2 = 154.5/18.5 = 8.35 or ω = 2.9 rad/s
∴ Velocity of striking = ω × OA = 2.9 (1.05 + 0.15) = 3.48 m/s Ans.
2. Impulse on the pendulum and sudden change of axis reaction
It is given that the impact value of the specimen (i.e. the energy used for breaking the specimen)
is 55 N-m. Let ω1 be the angular velocity of the pendulum immediately after impact. We know that
Loss of kinetic energy =
2 2 2 2 2
1 O 1
1 1( ) . ( ) 55
2 2
I m kω − ω = ω − ω = N-m
∴
2 2
1
1 30 1.233 (2.9 ) 55
2
× × − ω =
18.5 (8.41 – 2
1ω ) = 55 or 2
1ω = 8.41 – 55/18.5 = 5.44
∴ ω1 = 2.33 rad/s
Let P and Q be the impulses at the knife edge A and at the pivot O respectively as shown in
Fig. 4.8.
∴ P + Q = Change of linear momentum
= m.h (ω – ω1) = 30 × 1.05 (2.9 – 2.33) = 17.95 ... (i)
Taking moments about G,
0.15 P – 1.05 Q = Change of angular momentum
= 2 2 2
G 1 O 1. ( ) ( ) ( )m k m k hω − ω = − ω − ω
= 30 (1.233 – 1.052) (2.9 – 2.33) = 2.27 ... (ii)
From equations (i) and (ii),
P = 17.6 N-s; and Q = 0.35 N-s Ans.
∴ Change in axis reaction when pendulum is vertical
= Change in centrifugal force
2 2 2 2
1( ) 30 (2.9 2.33 )m h= ω − ω = − 1.05 = 94 N Ans.
4.10. Bifilar Suspension
The moment of inertia of a body may be determined experimentally by an apparatus called
bifilar suspension. The body whose moment of inertia is to be determined (say A B) is suspended by
two long parallel flexible strings as shown in Fig. 4.9. When the body is twisted through a small angle
θ about a vertical axis through the centre of gravity G, it will vibrate with simple harmonic motion in
a horizontal plane.
Let m = Mass of the body,
W = Weight of the body in newtons = m.g,
kG = Radius of gyration about an axis through the centre of gravity,
17. 88 l Theory of Machines
I = Mass moment of inertia of the
body about a vertical axis
through 2
G. ,G m k=
l = Length of each string,
x = Distance of A fromG (i.e. AG),
y = Distance of B from G (i.e. BG),
θ = Small angular displacement of
the body from the equilibrium
position in the horizontal plane,
φA and φB = Corresponding angular dis-
placements of the strings, and
α = Angular acceleration towards
the equilibrium position.
When the body is stationary, the tension in the strings are
given by
A B
. . . .
, and
m g y m g x
T T
x y x y
= =
+ + ...(Taking moments about B and A respectively,)
When the body is displaced from its equilibrium position in a horizontal plane through a
small angle θ, then the angular displacements of the strings are given by
A A′ = φA.l = x.θ ; and BB′ = φB.l = y.θ
∴ A B
..
; and
yx
l l
θθφ = φ =
Component of tensionTA in the horizontal plane, acting normal toA′B′ at A′as shown in Fig. 4.9
= A A
. . . . . ...
( )
m g y m g x yxT
x y l l x y
θθφ = × =
+ +
Component of tension TB in the horizontal plane, acting normal to A′B′ at B′as shown in Fig. 4.9
= B B
. . . . . . .
.
( )
m g x y m g x y
T
x y l l x y
θ θ
φ = × =
+ +
These components of tensions TA and TB are equal and opposite in direction, which gives rise
to a couple. The couple or torque applied to each string to restore the body to its initial equilibrium
position, i.e. restoring torque
A A B B. . . .T x T y= φ + φ
. . . . . . . .
( )
( )
m g x y m g x y
x y
l x y l
θ θ
= + =
+ ... (i)
and accelerating (or disturbing) torque
2
G. . .I m k= α = α ... (ii)
Equating equations (i) and (ii),
2
G
. . . .
. .
m g x y
mk
l
θ
= α or
2
G .
. .
k l
g x y
θ =
α
Fig. 4.9. Bifilar suspension.
18. Chapter 4 : Simple Harmonic Motion l 89
i.e.
2
G .Angular displacement
Angular acceleration . .
k l
g x y
=
We know that periodic time,
2
G.Angular displacement
2 2
Angular acceleration . .p
k l
t
g x y
= π = π
G2
. .
lk
g x y
= π
and frequency,
G
. .1 1
2p
g x y
n
t k l
= =
π
Note : The bifilar suspension is usually used for finding the moment of inertia of a connecting rod of an engine.
In this case, the wires are attached at equal distances from the centre of gravity of the connecting rod (i.e. x = y)
so that the tension in each wire is same.
Example 4.9. A small connecting rod of mass 1.5 kg is suspended in a horizontal plane by
two wires 1.25 m long. The wires are attached to the rod at points 120 mm on either side of the centre
of gravity. If the rod makes 20 oscillations in 40 seconds, find the radius of gyration and the mass
moment of inertia of the rod about a vertical axis through the centre of gravity.
Solution. Given : m = 1.5 kg ; l = 1.25 m ; x = y = 120 mm = 0.12 m
Since the rod makes 20 oscillations in 40 s, therefore frequency of oscillation,
n = 20/40 = 0.5 Hz
Radius of gyration of the connecting rod
Let kG = Radius of gyration of the connecting rod.
We know that frequency of oscillation (n),
G G
9.81 0.12 0.12. .1 1 0.05350.5
2 2 1.25
g x y
k l k k
× ×
= = =
π π
∴ kG = 0.0535/0.5 = 0.107 m = 107 mm Ans.
Mass moment of inertia of the connecting rod
We know that mass moment of inertia,
I = m (kG)2 = 1.5 (0.107)2 = 0.017 kg-m2 Ans.
4.11. Trifilar Suspension (Torsional Pendulum)
It is also used to find the moment of inertia of a body experi-
mentally. The body (say a disc or flywheel) whose moment of inertia
is to be determined is suspended by three long flexible wires A, B
and C, as shown in Fig. 4.10. When the body is twisted about its axis
through a small angle θ and then released, it will oscillate with simple
harmonic motion.
Let m = Mass of the body in kg,
W = Weight of the body in newtons = m.g,
kG = Radius of gyration about an axis
through c.g.,
I = Mass moment of inertia of the disc about an axis through O and per-
pendicular to it = m.k2,
Fig. 4.10. Trifilar suspension.
19. 90 l Theory of Machines
l = Length of each wire,
r = Distance of each wire from the axis of the disc,
θ = Small angular displacement of the disc,
φ = Corresponding angular displacement of the wires, and
α = Angular acceleration towards the equilibrium position.
Then, for small displacements,
r. θ = l. φ or φ = r.θ/l
Since the three wires are attached symmetrically with respect to the axis, therefore the ten-
sion in each wire will be one-third of the weight of the body.
∴ Tension in each wire = m.g/3
Component of the tension in each wire perpendicular to r
. .sin . . . . .
3 3 3
m g m g m g r
l
φ φ θ
= = = . . . ( ∵ φ is a small angle, and φ = r.θ/l)
∴ Torque applied to each wire to restore the body to its initial equilibrium position i.e.
restoring torque
2
. . . . . .
3 3
m g r m g r
r
l l
θ θ
= × =
Total restoring torque applied to three wires,
2 2
. . . . . .
3
3
m g r m g r
T
l l
θ θ
= × = ... (i)
We know that disturbing torque
= I.α = 2
G. .m k α ... (ii)
Equating equations (i) and (ii),
2
2
G
. . .
. .
m g r
m k
l
θ
= α or
2
G
2
.
.
l k
g r
θ
=
α
i.e.
2
G
2
.Angular displacement
Angular acceleration .
l k
g r
=
We know that periodic time,
2
G G
2
. 2Angular displacement
2 2
Angularacceleration .
p
l k k lt
r gg r
π
= π = π =
and frequency,
G
1
2p
grn
t k l
= =
π
Example 4.10. In order to find the radius of gyration of a car, it is suspended with its axis
vertical from three parallel wires 2.5 metres long. The wires are attached to the rim at points spaced
120° apart and at equal distances 250 mm from the axis.
It is found that the wheel makes 50 torsional oscillations of small amplitude about its axis in
170 seconds. Find the radius of gyration of the wheel.
Solution. Given : l = 2.5 m ; r = 250 mm = 0.25 m ;
Since the wheel makes 50 torsional oscillations in 170 seconds, therefore frequency of
oscillation,
n = 50/170 = 5/17 Hz
Let kG = Radius of gyration of the wheel
20. Chapter 4 : Simple Harmonic Motion l 91
We know that frequency of oscillation (n),
G G G
5 0.25 9.81 0.079
17 2 2 2.5
r g
k l k k
= = =
π π
∴ kG = 0.079 × 17/5 = 0.268 m = 268 mm Ans.
Example 4.11. A connecting rod of mass 5.5 kg is placed on a horizontal platform whose
mass is 1.5 kg. It is suspended by three equal wires, each 1.25 m long, from a rigid support. The wires
are equally spaced round the circumference of a circle of 125 mm radius. When the c.g. of the
connecting rod coincides with the axis of the circle, the platform makes 10 angular oscillations in 30
seconds. Determine the mass moment of inertia about an axis through its c.g.
Solution. Given : m1 = 5.5 kg ; m2 = 1.5 kg ; l = 1.25 m ; r = 125 mm = 0.125 m
Since the platform makes 10 angular oscillations in 30 s, therefore frequency of oscillation,
n = 10/30 = 1/3 Hz
Let kG = Radius of gyration about an axis through the c.g.
We know that frequency of oscillation (n)
G G G
1 0.125 9.81 0.056
3 2 2 1.25
gr
k l k k
= = =
π π
∴ kG = 0.056 × 3 = 0.168 m
and mass moment of inertia about an axis through its c.g.,
2 2 2 2
G 1 2 G. ( ) (5.5 1.5) (0.168) kg-mI m k m m k= = + = +
= 0.198 kg-m2 Ans.
EXERCISES
1. A particle, moving with simple harmonic motion, performs 10 complete oscillations per minute and
its speed, when at a distance of 80 mm from the centre of oscillation is 3/5 of the maximum speed.
Find the amplitude, the maximum acceleration and the speed of the particle, when it is 60 mm from
the centre of the oscillation. [Ans. 100 mm ; 109.6 mm/s2 ; 83.76 mm/s]
2. A piston, moving with a simple harmonic motion, has a velocity of 8 m/s, when it is 1 metre from the
centre position and a velocity of 4 m/s, when it is 2 metres from the centre. Find : 1. Amplitude, 2.
Periodic time, 3. Maximum velocity, and 4. Maximum acceleration.
[Ans. 2.236 m ; 1.571 s ; 8.94 m/s ; 35.77 m/s2]
3. The plunger of a reciprocating pump is driven by a crank of radius 250 mm rotating at 12.5 rad/s.
Assuming simple harmonic motion, determine the maximum velocity and maximum acceleration of
the plunger. [Ans. 3.125 m/s ; 39.1 m/s2]
4. A part of a machine of mass 4.54 kg has a reciprocating motion which is simple harmonic in character.
It makes 200 complete oscillations in 1 minute. Find : 1. the accelerating force upon it and its velocity
when it is 75 mm, from midstroke ; 2. the maximum accelerating force, and 3. the maximum velocity
if its total stroke is 225 mm i.e. if the amplitude of vibration is 112.5 mm.
[Ans. 149.5 N ; 1.76 m/s ; 224 N ; 2.36 m/s]
5. A helical spring of negligible mass is required to support a mass of 50 kg. The stiffness of the spring
is 60 kN/m. The spring and the mass system is displaced vertically by 20 mm below the equilibrium
position and then released. Find : 1. the frequency of natural vibration of the system ; 2. the velocity
and acceleration of the mass when it is 10 mm below the rest position.
[Ans. 5.5 Hz ; 0.6 m/s ; 11.95 m/s2]
6. A spring of stiffness 2 kN/m is suspended vertically and two equal masses of 4 kg each are attached to
the lower end. One of these masses is suddenly removed and the system oscillates. Determine : 1. the
amplitude of vibration, 2. the frequency of vibration, 3. the velocity and acceleration of the mass when
21. 92 l Theory of Machines
passing through half amplitude position, and 4. kinetic energy of the vibration in joules.
[Ans. 0.019 62 m ; 3.56 Hz ; 0.38 m/s , 4.9 m/s2 ; 0.385 J]
7. A vertical helical spring having a stiffness of 1540 N/m is clamped at its upper end and carries a mass
of 20 kg attached to the lower end. The mass is displaced vertically through a distance of 120 mm and
released. Find : 1. Frequency of oscillation ; 2. Maximum velocity reached ; 3. Maximum accelera-
tion; and 4. Maximum value of the inertia force on the mass.
[Ans. 1.396 Hz ; 1.053 m/s ; 9.24 m/s2 ; 184.8 N]
8. A small flywheel having mass 90 kg is suspended in a vertical plane as a compound pendulum. The
distance of centre of gravity from the knife edge support is 250 mm and the flywheel makes 50
oscillations in 64 seconds. Find the moment of inertia of the flywheel about an axis through the centre
of gravity. [Ans. 3.6 kg-m2]
9. The connecting rod of a petrol engine has a mass 12 kg. In order to find its moment of inertia it is
suspended from a horizontal edge, which passes through small end and coincides with the small end
centre. It is made to swing in a vertical plane, such that it makes 100 oscillations in 96 seconds. If the
point of suspension of the connecting rod is 170 mm from its c.g., find : 1. radius of gyration about an
axis through its c.g., 2. moment of inertia about an axis through its c.g., and 3. length of the equivalent
simple pendulum. [Ans. 101 mm ; 0.1224 kg-m2 ; 0.23 m]
10. A connecting rod of mass 40 kg is suspended vertically as a compound pendulum. The distance between
the bearing centres is 800 mm. The time for 60 oscillations is found to be 92.5 seconds when the axis of
oscillation coincides with the small end centre and 88.4 seconds when it coincides with the big end
centre. Find the distance of the centre of gravity from the small end centre, and the moment of inertia of
the rod about an axis through the centre of gravity. [Ans. 0.442 m ; 2.6 kg-m2]
11. The following data were obtained from an experiment to find the moment of inertia of a pulley by
bifilar suspension :
Mass of the pulley = 12 kg ; Length of strings = 3 m ; Distance of strings on either side of centre of
gravity = 150 mm ; Time for 20 oscillations about the vertical axis through c.g. = 46.8 seconds
Calculate the moment of inertia of the pulley about the axis of rotation.
[Ans. 0.1226 kg-m2]
12. In order to find the moment of inertia of a flywheel, it is suspended in the horizontal plane by three
wires of length 1.8 m equally spaced around a circle of 185 mm diameter. The time for 25 oscillations
in a horizontal plane about a vertical axis through the centre of flywheel is 54 s. Find the radius of
gyration and the moment of inertia of the flywheel if it has a mass of 50 kg.
[Ans. 74.2 mm; 0.275 kg-m2]
DO YOU KNOW ?
1. Explain the meaning of S.H.M. and give an example of S.H.M.
2. Define the terms amplitude, periodic time, and frequency as applied to S.H.M.
3. Show that when a particle moves with simple harmonic motion, its time for a complete
oscillation is independent of the amplitude of its motion.
4. Derive an expression for the period of oscillation of a mass when attached to a helical spring.
5. What is a simple pendulum ? Under what conditions its motion is regarded as simple harmonic?
6. Prove the formula for the frequency of oscillation of a compound pendulum. What is the length of a
simple pendulum which gives the same frequency as compound pendulum ?
7. Show that the minimum periodic time of a compound pendulum is
G
( )
2
2p min
k
t
g
= π
where kG is the radius of gyration about the centre of gravity.
8. What do you understand by centre of percussion ? Prove that it lies below the centre of gravity of the
body and at a distance 2
G /k h , where kG is the radius of gyration about c.g. and h is the distance
between the centre of suspension and centre of gravity.
22. Chapter 4 : Simple Harmonic Motion l 93
9. Describe the method of finding the moment of inertia of a connecting rod by means of bifilar suspen-
sion. Derive the relations for the periodic time and frequency of oscillation.
10. What is a torsional pendulum ? Show that periodic time of a torsional pendulum is
G2
p
k lt
r g
π
=
where kG = Radius of gyration,
l = Length of each wire, and
r = Distance of each wire from the axis of the disc.
OBJECTIVE TYPE QUESTIONS
1. The periodic time (tp) is given by
(a) ω / 2π (b) 2 π / ω (c) 2 π × ω (d) π/ω
2. The velocity of a particle moving with simple harmonic motion is . . . . at the mean position.
(a) zero (b) minimum (c) maximum
3. The velocity of a particle (v) moving with simple harmonic motion, at any instant is given by
(a) 2 2
r xω − (b) 2 2
x rω − (c) 2 2 2
r xω − (d) 2 2 2
x rω −
4. The maximum acceleration of a particle moving with simple harmonic motion is
(a) ω (b) ω.r (c) ω2.r (d) ω2/r
5. The frequency of oscillation for the simple pendulum is
(a)
1
2
L
gπ
(b)
1
2
g
Lπ
(c) 2
L
g
π (d) 2
g
L
π
6. When a rigid body is suspended vertically and it oscillates with a small amplitude under the action of the
force of gravity, the body is known as
(a) simple pendulum (b) torsional pendulum
(c) compound pendulum (d) second’s pendulum
7. The frequency of oscillation of a compound pendulum is
(a) 2 2
G
.1
2
g h
k hπ +
(b)
2 2
G1
2 .
k h
g h
+
π
(c)
2 2
G
.
2
g h
k h
π
+
(d)
2 2
G
2
.
k h
g h
+
π
where kG = Radius of gyration about the centroidal axis, and
h = Distance between the point of suspension and centre of gravity of the body.
8. The equivalent length of a simple pendulum which gives the same frequency as the compound pendulum
is
(a) 2 2
G
h
k h+
(b)
2 2
Gk h
h
+
(c)
2
2 2
G
h
k h+
(d)
2 2
G
2
k h
h
+
9. The centre of percussion is below the centre of gravity of the body and is at a distance equal to
(a) h / kG (b) h.kG (c) h2/kG (d) 2
G /k h
10. The frequency of oscillation of a torsional pendulum is
(a) G2 k g
r l
π
(b)
G2
gr
k lπ
(c)
G2 k l
r g
π
(d)
G2
r l
k gπ
ANSWERS
1. (b) 2. (c) 3. (a) 4. (c) 5. (b)
6. (c) 7. (a) 8. (b) 9. (d) 10. (b)
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