Learning object 1 (shm)
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A clicker-type question that tests students' knowledge on interpreting graphs of simple harmonic motion.
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Max Yuen
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(use g = 9.8m/s2 for all problems.)
Background
Many physics problems cannot be solved directly by hand or analytically. We resort to numerical
methods to give us approximations to the problem. In this lab you will learn the Euler method,
which allows you to solve Newton’s laws of motion. This is done by treating the velocity as a
piecewise linear function with many time intervals and during interval the acceleration is assumed
to be uniform. This allows us to use the kinematic equations we learned about in the first half
of the class to approximate the motion. If we choose to partition the motion into smaller time
intervals, the approximation becomes much better since the differences between adjacent intervals
become smaller. In this lab, this numerical analysis method will be applied to the motion of a
falling object under the influence of gravity and drag force. If you are adventurous, you can even
try to extend this to 2D and compute the realistic trajectory of a baseball. You might even try
some other problems, such as a mass attached to a spring.
Euler’s Method Foundations
This method is well suited for problems where the acceleration is a function of the velocity, as in
the case of a falling object under the influence of gravity and drag:
a = f(v) (1)
Falling object with drag force
The model for drag fits the prescription for using Euler’s method since the net force on a falling
object with drag is given by:
ma = −mg −FD (2)
ma = −mg −
1
2
ρairACDv
2 · sgn(v) (3)
a = −g
(
1 +
ρairACDv
2 · sgn(v)
2mg
)
(4)
a = f(v) ← Equation of Motion (5)
where m is the mass of the falling object, a is the acceleration of the object (which is positive when
pointed up), ρair is the density of air (about 1.29 ·10−3kg/m3), A is the cross-sectional area, CD is
the drag coefficient, v is the object’s velocity, and sgn(v) is the signum function which returns the
sign of the argument. The second signum function is there to guarantee that the direction of the
drag force is always in the opposite direction of the velocity function. Note that we see that the
acceleration is an explicit function of v, which sort of makes this a chicken or egg problem. This is
because we need a to get v, but to get a we need v, so which one do we compute first? Hold that
thought. We’ll talk more on how to program this in EXCEL or Google Sheets later.
1
Figure 1: FBD for an object falling under the pull of gravity and resistance by drag force
Terminal Velocity
In lecture, we talked about how after waiting for some time, if the object started at rest the
speed will increase and the drag force will also become larger and eventually balance out with the
gravitational force. When this happens, we have reached terminal velocity vterm = −v. This can
be solved by setting a = 0:
0 = −mg −
1
2
ρairACDv
2 · sgn(v) (6)
2mg = ρairACDv
2
term (7)
→ vterm =
√
2mg
ρairACD
(8)
Using this definition for the terminal ...
Mit2 092 f09_lec20
All of material inside is un-licence, kindly use it for educational only but please do not to commercialize it.
Based on 'ilman nafi'an, hopefully this file beneficially for you.
Thank you.
03 time and motion
The document discusses generating smooth trajectories for moving objects from an initial pose to a final pose over time. It describes how to create single-dimensional and multi-dimensional trajectories using polynomial and trapezoidal functions. It also covers generating multi-segment trajectories to smoothly move through via points without stopping by using polynomial blends between linear motion segments.
Introduction to basic principles of fluid mechanics
1) The document introduces basic principles of fluid mechanics, including Lagrangian and Eulerian descriptions of fluid flow. The Lagrangian description follows individual particles, while the Eulerian description observes flow properties at fixed points in space.
2) It describes three governing laws of fluid motion within a control volume: conservation of mass (the net flow in and out of a control volume is zero), conservation of momentum (Newton's second law applied to a fluid system), and conservation of energy.
3) It derives Bernoulli's equation, which relates pressure, velocity, and elevation along a streamline for inviscid, steady, incompressible flow. Bernoulli's equation is an application of conservation of momentum along a streamline.
4.1
Oscillations and waves can be described by key parameters including amplitude, period, and frequency. Amplitude refers to the maximum displacement from equilibrium, period is the time for one full oscillation, and frequency is the number of oscillations per second. Common examples of oscillations include a pendulum, mass on a spring, and ocean tides. For an object to undergo simple harmonic motion, it must experience a restoring force proportional to and directed towards its displacement from equilibrium.
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simple pendulum
This question is the review for energy, and also combines the idea of simple pendulum which is good to be used to understand the motion of the pendulum as well as the equation for pendulum
Simple pendulum
This question is the review for energy, and also combines the idea of simple pendulum which is good to be used to understand the motion of the pendulum as well as the equation for pendulum
Simple Pendulum
The document describes a simple pendulum experiment with a 0.1kg mass attached to a 0.4m string. It asks three questions: (1) What is the tension force on the pendulum when it is at point B at an angle of 30 degrees? (2) What is the speed of the pendulum when it crosses point C? (3) What is the speed of the pendulum when it is at point B? The answers provided are: (1) 0.85N, (2) 2.8m/s, (3) 2.6m/s.
impuls and momentum
This document discusses momentum and collisions. It defines momentum as the product of an object's mass and velocity. It explains that momentum is conserved in collisions according to the law of conservation of momentum. It also discusses different types of collisions, including perfectly elastic collisions where both momentum and kinetic energy are conserved, and inelastic collisions where kinetic energy is not conserved. Examples of applications to rockets and collisions are provided. Learning activities and assessments are outlined to help students understand these concepts.
Physics 101 Learning Object #1
This document contains a physics learning object and solutions for simple harmonic motion. It includes graphs of displacement, velocity and acceleration over time. Students are asked to find the amplitude and angular frequency from the graphs. They are also asked which statement about simple harmonic motion is false - the answer is that the displacement and acceleration of an oscillator are always related, not unrelated as stated.
Modification factor
The document discusses modification factors, which describe how a system's steady-state response is magnified under harmonic loading compared to its complete response including transients. While the steady-state response has an analytical solution, the complete response does not have a simple closed-form expression. The document presents an analytical expression for the peak of the complete undamped response and compares it to the steady-state response amplitude. Results are discussed as a function of the ratio of the excitation frequency to the undamped natural frequency of the system.
Simple harmonic motion
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
From the Front LinesOur robotic equipment and its maintenanc.docx
From the Front Lines
Our robotic equipment and its maintenance represent a fixed cost of $23,320 per month. The cost-effectiveness of robotic-assisted surgery is related to patient volume: With only 10 cases, the fixed cost per case is $2,332, and with 40 cases, the fixed cost per case is $583.
Source: Alemozaffar, Chang, Kacker, Sun, DeWolf, & Wagner (2013).
MATLAB sessions: Laboratory 5
MAT 275 Laboratory 5
The Mass-Spring System
In this laboratory we will examine harmonic oscillation. We will model the motion of a mass-spring
system with differential equations.
Our objectives are as follows:
1. Determine the effect of parameters on the solutions of differential equations.
2. Determine the behavior of the mass-spring system from the graph of the solution.
3. Determine the effect of the parameters on the behavior of the mass-spring.
The primary MATLAB command used is the ode45 function.
Mass-Spring System without Damping
The motion of a mass suspended to a vertical spring can be described as follows. When the spring is
not loaded it has length ℓ0 (situation (a)). When a mass m is attached to its lower end it has length ℓ
(situation (b)). From the first principle of mechanics we then obtain
mg︸︷︷︸
downward weight force
+ −k(ℓ − ℓ0)︸ ︷︷ ︸
upward tension force
= 0. (L5.1)
The term g measures the gravitational acceleration (g ≃ 9.8m/s2 ≃ 32ft/s2). The quantity k is a spring
constant measuring its stiffness. We now pull downwards on the mass by an amount y and let the mass
go (situation (c)). We expect the mass to oscillate around the position y = 0. The second principle of
mechanics yields
mg︸︷︷︸
weight
+ −k(ℓ + y − ℓ0)︸ ︷︷ ︸
upward tension force
= m
d2(ℓ + y)
dt2︸ ︷︷ ︸
acceleration of mass
, i.e., m
d2y
dt2
+ ky = 0 (L5.2)
using (L5.1). This ODE is second-order.
(a) (b) (c) (d)
y
ℓ
ℓ0
m
k
γ
Equation (L5.2) is rewritten
d2y
dt2
+ ω20y = 0 (L5.3)
c⃝2011 Stefania Tracogna, SoMSS, ASU
MATLAB sessions: Laboratory 5
where ω20 = k/m. Equation (L5.3) models simple harmonic motion. A numerical solution with ini-
tial conditions y(0) = 0.1 meter and y′(0) = 0 (i.e., the mass is initially stretched downward 10cms
and released, see setting (c) in figure) is obtained by first reducing the ODE to first-order ODEs (see
Laboratory 4).
Let v = y′. Then v′ = y′′ = −ω20y = −4y. Also v(0) = y′(0) = 0. The following MATLAB program
implements the problem (with ω0 = 2).
function LAB05ex1
m = 1; % mass [kg]
k = 4; % spring constant [N/m]
omega0 = sqrt(k/m);
y0 = 0.1; v0 = 0; % initial conditions
[t,Y] = ode45(@f,[0,10],[y0,v0],[],omega0); % solve for 0<t<10
y = Y(:,1); v = Y(:,2); % retrieve y, v from Y
figure(1); plot(t,y,’b+-’,t,v,’ro-’); % time series for y and v
grid on;
%-----------------------------------------
function dYdt = f(t,Y,omega0)
y = Y(1); v = Y(2);
dYdt = [ v ; -omega0^2*y ];
Note that the parameter ω0 was passed as an argument to ode45 rather than set to its value ω0 = 2
directly in the funct ...
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Learning object 1 (shm)
- 1. Simple Harmonic Motion Learning Object By Eugenie Kwong PHYSICS 101-202 Group LC2
- 2. Given the following position-time graph of a simple harmonic oscillator, at what times will the velocity be at its maximum? What times will the acceleration be at its maximum? A) Maximum velocity at t=0.5 + 0.5T, maximum acceleration at t=0 + 0.5T B) Maximum velocity at t=0.5 + T, maximum acceleration at t=0+T C) Maximum velocity at t=0 + 0.5T, maximum acceleration at t=0.5+0.5T D) Maximum velocity at t = 0 + T, maximum acceleration at t=0.5+T E) Both the maximum velocity and acceleration occur at t=0+0.5T Note: t = time, T = period http://www.unistudyguides.com/wiki/File:SHM_2.jpg
- 3. Solution: A) Maximum velocity at t=0.5 + 0.5T, maximum acceleration at t=0 + 0.5T One can think about the conservation of energy. When the position is 0m, the potential energy is also 0 and so the only energy present is kinetic energy – resulting in a maximum velocity at this point. This repeats for every half period (answer B is not fully correct as it only accounts for either the +v or –v; it should include both). The maximum acceleration occurs when the displacement is also at its maximum values. “For a motion to be simple harmonic motion, the acceleration of the object must be proportional to its displacement from the equilibrium position and opposite in direction, at all times.” http://www.unistudyguides.com/wiki/File:SHM_2.jpg
- 4. Answer C is incorrect because at t=0+0.5T, the displacement is at its maximum. Again with conservation of energy, the total energy at that point is equal to only the potential energy (no kinetic energy). For this reason, the velocity would be 0m/s at such times. Answer D is incorrect, following the same reason explained for answer C. Answer E is incorrect. Although the description for acceleration is correct, the times for velocity is incorrect (see reasoning for answer C). Explanation continued…