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.

1. Simple Harmonic Motion.

2. Revision. For this part of the course you need to be familiar with the concepts of circular motion and angular velocity. State the formulas used to calculate angular velocity.

3. SHM A system will oscillate if there is a force acting on it that tends to pull it back to its equilibrium position – a restoring force. In a swinging pendulum the combination of gravity and the tension in the string that always act to bring the pendulum back to the centre of its swing. Tension Gravity Resultant (restoring force)

4. For a spring and mass the combination of the gravity acting on the mass and the tension in the spring means that the system will always try to return to its equilibrium position. M T g Restoring force

5. If the restoring force f is directly proportional to the displacement x , the oscillation is known as simple harmonic motion (SHM). For an object oscillating with SHM f  -x The minus sign shows that the restoring force is acting opposite to the displacement. Acceleration In SHM

6. Displacement Restoring force

7. Kinetic and Potential energy in SHM

8. We already know that f=ma, we can substitute this into the above formula to give. a  -x If we put in a constant we get the equation – a = -  2 x  2 is a constant,  is called the angular velocity and is dependant on the frequency of the oscillations and can be written as  =2  f. So the equation for SHM can be written as: a = -(2  f) 2 x

9. Questions. Use the equation a = -  2x to work out the correct units of  . Sketch an oscillating pendulum and mark in the positions of where the acceleration is greatest and smallest. An ultrasonic welder uses a tip that vibrates at 25 kHz if the tip’s amplitude is 6.25x10-2 calculate the maximum acceleration of the tip

10. A wave has a frequency of 4 Hz and an amplitude of 0.3m calculate its maximum acceleration. A system is oscillating at 300 kHz with an amplitude of 0.6 mm calculate its maximum acceleration. A speaker cone playing a constant bass note is oscillating at 100 Hz, the total movement of the speaker cone is 1 cm. Assuming that the movement is SHM calculate its maximum acceleration.

11. Displacement. If a system oscillates with SHM the pattern of the motion will be the same, the motion will follow the same rules. The equation for the displacement, x , is related to the time, t , by the equation: x = Acos  t (remember to use radians)

12. We already know that  can also be written as 2  f, then the equation can also be written as: x = Acos2  ft We can use this equation to predict the position of an oscillating system at any time.

13. Qs A pendulum is oscillating at 30 times per minute and has an amplitude of 20 cm, find its amplitude 0.5s after being released from its maximum displacement. Find its displacement 0.75s after being released from its maximum displacement. If the pendulum is oscillating at 120 times per minute and has an amplitude of 30 cm. Work out the displacement at 0.2 s, 0.4 s and 0.5 s

14. If high tide is at 12 noon and the next is 12 hours later and the amplitude of the tide is 2m we can work out the height of the tide at any time for example 2pm. Find out the height of the tide 2pm later (4pm).

15. Energy in simple harmonic motion. We already know that the total energy (mechanical energy) for a system that is moving with SHM is the sum of the potential and kinetic energies. When the object is at the extremes of its oscillation (x =  A) it has no KE but the PE is at its maximum. When the object is mid way through its oscillation (when x = 0) the KE is at its maximum but there is no PE.

16. The potential energy is equal to the work done. Ep = ½ m  2 x 2 At the maximum displacement E = Ep so E = ½ m  2 A 2 And ½ m  2 A 2 = ½ m  2 x 2 + ½ mv 2 If we divide ½ m.  2 A 2 =  2 x 2 + v 2

17. So v 2 =  2 (A 2 - x 2 ) or v =  2  f  A 2 -x 2 We can use this equation to workout the velocity of an object at any position in it’s SHM.

18. A 1 kg mass is hanging from a spring which has a spring constant (k) of 1000 Nm-1, it is oscillating with an amplitude of 2cm. a) Use T = 2   m/k to calculate the time period of the oscillation. b) use v=  2  f  A 2 -x 2 to find the velocity and then the kinetic energy of the mass. -0.02 etc 0.01 0.015 0.02 KE, (joules) Velocity, (m/s) Displacement, x (m)

19. Qs A metal strip is clamped to the edge of the table and has an object of mass 280g attached to the free end. The object is pulled down and released. The object vibrates with SHM with an amplitude of 8.0 cm and a period of 0.16 s. Calculate the maximum acceleration of the object Calculate the maximum force State the position of the object when it has no KE.

20. Describing SHM

21. http://physics.bu.edu/~duffy/semester1/c18_SHM_graphs.html

22. Displacement (x) Acceleration (a)  out of phase (180 deg) Velocity (v)  /2 out of phase (90 deg)

23. Simple pendulum A pendulum consists of a small “bob” of mass m , suspended by a light inextensible thread of length l , from a fixed point. The bob can be made to oscillate about point O in a vertical plane along the arc of a circle. We can ignore the mass of the thread

24. We can show that oscillating simple pendulums exhibit SHM. We need to show that a  x. Consider the forces acting on the pendulum: weight , W of the bob and the tension , T in the thread. We can resolve W into 2 components parallel and perpendicular to the thread: Parallel: the forces are in equilibrium Perpendicular: only one force acts, providing acceleration back towards O

26. Parallel to string: F = mg cos  Perpendicular: to string F = restoring force towards O = mg sin  We already know that F = ma So F= -mg sin  = ma (-ve since towards O)

27. At small angles (  = less than 10 deg) Sin  is approximately equal to  (in radians)  approximates to x/l for small angles. So:  -mg ( x /l) = ma Rearranging: a = -g ( x /l) = pendulum equation (can also write this equation as a = - x (g/l) )

28. In SHM a  x SHM equation a = -(2  f )2 x Pendulum equation a = - x (g/l) Hence (2  f ) 2 = (g/l)  f = 1/2  (  g/l) (remember T = 1/ f) T = 2  (  l/g) The time period of a simple pendulum depends on length of thread and acceleration due to gravity

29. Experiment Using a long clamp stand, a pendulum bob, some light string and a stop watch to investigate the relation ship between g, l and T For a pendulum of known length count the time taken for 10 complete oscillations (there and back). Use the pendulum formula to calculate the force of gravity. Repeat with 3 other lengths.

30. SHM in Springs

31. In a spring-mass system. Do you think the size of the mass affects the Time Period of the Oscillation? What do you think the relationship will be? Do you think the stiffness (spring constant) of the spring will affect the Time Period of the Oscillation? What do you think the relationship will be?

32. The diagram here shows a mass-spring system. Set the equipment up and use F=Ke to determine the spring constant of the spring. Use 3 different masses to determine the time period of oscillation. Double up the springs in parallel and series and try to determine the period of oscillation. M

33. How did the different masses affect the period of oscillation. How did the spring constant (different arrangements of springs) affect the period of oscillation?

34. The time period of oscillation of a spring is dependant on the spring constant of the spring and the mass of the system. It is independent of the force of gravity . The relationship is.

35. Q1 Calculate the time period of a spring mass system of 2.5 kg with a spring const. of 200N/m. Q2 What is the frequency of a 20g mass oscillating on the end of a spring with a const. of 120 N/m. Q3 A spring is oscillating 45 times per min. calculate the mass if the const. is 1000 N/m. Q4 If a mass of 10000 Kg is oscillating at a frequency of 0.37 Hz what is the const. of the spring?

36. Resonance and damping.

37. Resonance. Resonance is the tendency in a system to vibrate at its maximum amplitude at a certain frequency. This frequency is known as the system's resonance frequency. When damping is small, the resonance frequency is approximately equal to the natural frequency of the system, which is the frequency of free vibrations. The natural or fundamental frequency is often written as f 0

38. Perhaps one of the more common examples of resonance is in musical instruments. For example in guitars it is possible to make other strings vibrate “sympathetically” when another is plucked, either at their fundamental or overtone frequencies. http://www.youtube.com/watch?v=MBZs5SCtlVA

39. Examples of resonance. Pushing a child on a swing – maximum A when pushing  =  o Tuning a radio – electrical resonance occurs when  o of tuning circuit adjusted to match  of incoming signal Pipe instruments - column of air forced to vibrate. If reed  =  o of column loud sound produced Rotating machinery – e.g. washing machine. An out of balance drum will result in violent vibrations at certain speeds

40. Tacoma narrows bridge . The Tacoma narrows bridge is often used as an example of resonance, although it is not strictly scientifically accurate to do so. It does how ever give an example of what can happen if an object was to be kept at its resonant frequency for a long time. http://www.youtube.com/watch?v=j-zczJXSxnw&feature=related

41. Barton’s Pendulum All objects have a natural frequency of vibration or resonant frequency. If you force a system - in this case a set of pendulums - to oscillate, you get a maximum transfer of energy, i.e. maximum amplitude imparted. When the driving frequency equals the resonant frequency of the driven system. The phase relationship between the driver and driven oscillator is also related by their relative frequencies of oscillation.

43. You also get a very clear illustration of the phase of oscillation relative to the driver. The pendulum at resonance is π/2 behind the driver, all the shorter pendulums are in phase with the driver and all the longer ones are π out of phase. The amplitude of the forced oscillations depend on the forcing frequency of the driver and reach a maximum when forcing frequency = natural frequency of the driven cones.

44. Another example of resonance in a driven system is the hacksaw blade oscillator. Driving mass And arm ‘Slave’ arm with ‘slave’ mass Elastic band pointer

45. If we change the period of oscillation of the driver by moving the mass (increasing L) the hacksaw blade will vibrate at different rates, if we get the driving frequency right the slave will reach resonant frequency and vibrate wildly. If we move the masses on the blade it will have a similar effect.

46. Problems with resonance. Resonance driver applies forces that continually supply energy to oscillator  increasing amplitude. A increases indefinitely unless energy transferred away. Severe case: A limit reached when oscillator destroys itself. E.g. wine glass shatters when opera singer reaches particular note. How do we deal with unwanted resonance? We could use damping, we could also change  o of object by changing its mass, if we were to change the stiffness of supports (  moving resonant  away from driving  ) we could reduce the affect of resonance.

47. Damping The amplitude depends on the degree of damping

48. A damped spring Set up a suspended mass-spring system with a ‘damper’ – a piece of card attached horizontally to the mass to increase the air drag. Alternatively, clamp a springy metal blade (e.g. hacksaw blade) firmly to the bench. Attach a mass to the free end, and add a damping card. Show how the amplitude decreases with time.