1. 4.1c Further Mechanics SHM & Oscillations Breithaupt pages 34 to 49 January 4 th 2012

2. AQA A2 Specification LessonsTopics 1 to 3Simple harmonic motion Characteristic features of simple harmonic motion. Condition for shm: a = − (2πf ) 2 x X = A cos 2πf t and v = ± 2πf √(A 2 − x 2 ) Graphical representations linking x, v, a and t. Velocity as gradient of displacement-time graph. Maximum speed = 2πfA. Maximum acceleration = (2πf ) 2 A. 4 to 6Simple harmonic systems Study of mass-spring system. T = 2π√(m / k) Study of simple pendulum. T = 2π√(l / g) Variation of Ek, Ep and total energy with displacement, and with time. 7 to 9Forced vibrations and resonance Qualitative treatment of free and forced vibrations. Resonance and the effects of damping on the sharpness of resonance. Phase difference between driver and driven displacements. Examples of these effects in mechanical systems and stationary wave situations

3. Oscillations (definitions) An oscillation is a repeated motion about a fixed point. The fixed point, known as the equilibrium position, is where the oscillating object returns to once the oscillation stops. The time period, T of an oscillation is the time taken for an object to perform one complete oscillation. Time = 0 periodTime = T / 8Time = T / 4Time = 3T / 8Time = T / 2Time = 5T / 8Time = 3T / 4Time = 7T / 8Time = 1 period equilibrium position

4. The amplitude, A of an oscillation is equal to the maximum value of the displacement, x Frequency, f in hertz is equal to the number of complete oscillations per second. also: f = 1 / T Angular frequency, ω in radians per second is given by: ω = 2π f or ω = 2π / T equilibrium position x = 0x = + Ax = 0x = - Ax = 0

5. Simple Harmonic Motion Many oscillating systems undergo a pattern of oscillation that is, or approximately the same as, that known as Simple Harmonic Motion (SHM). Examples (including some that are approximate): A mass hanging from the end of a spring Molecular oscillations A pendulum or swing A ruler oscillating on the end of a bench The oscillation of a guitar or violin string Tides Breathing

6. The pattern of SHM motion The pattern of SHM is the same as the side view of an object moving at a constant speed around a circular path The amplitude of the oscillation is equal to the radius of the circle. The object moves quickest as it passes through the central equilibrium position. The time period of the oscillation is equal to the time taken for the object to complete the circular path. + A - A

7. Conditions required for SHM When an object is performing SHM: 1. Its acceleration is proportional to its displacement from the equilibrium position. 2. Its acceleration is directed towards the equilibrium position. Mathematically the above can be written: a = - k x where k is a constant and the minus sign indicates that the acceleration, a and displacement, x are in opposite directions

8. Acceleration variation of SHM The constant k is equal to (2 π f ) 2 or ω 2 or (2 π /T) 2 Therefore: a = - (2πf ) 2 x (given on data sheet) or a = - ω 2 x or a = - (2π/T) 2 x x a + A- A - ω 2 x + ω 2 x gradient = - ω 2 or - (2πf ) 2

9. Question A body oscillating with SHM has a period of 1.5s and amplitude of 5cm. Calculate its frequency and maximum acceleration. (a) f = 1 / T = 1 / 1.5s frequency = 0.67 Hz (b) a = - (2πf ) 2 x maximum acceleration is when x = A (the amplitude) a = - (2πf ) 2 A = - (2π x 0.6667Hz) 2 x 0.015m maximum acceleration = 0.88 ms -2

10. Displacement equations of SHM The displacement, x of the oscillating object varies with time, t according to the equations: x = A cos (2π f t) given on data sheet or x = A cos (ω t) or x = A cos ((2π / T) t) Note: At time, t = 0, x = +A

11. Question A body oscillating with SHM has a frequency of 50Hz and amplitude of 4.0mm. Calculate its displacement and acceleration 2.0ms after it reaches its maximum displacement. Maximum displacement = amplitude, A = 4.0mm (a) x = A cos (2π f t) = 4.0mm x cos (2π 50 x 2.0ms) = 4.0mm x cos (2π 50 x 0.0020s) = 0.6283 displacement = 0.628 mm (b) a = - (2πf ) 2 x = - (2π x 50) 2 x 0.6283mm = - (100π) 2 x 0.0006283m acceleration = - 62 ms -2

12. Velocity equations of SHM The velocity, v of an object oscillating with SHM varies with displacement, x according to the equations: v = ± 2πf √(A 2 − x 2 ) given on data sheet or v = ± ω√(A 2 − x 2 ) or v = ± (2π / T) √(A 2 − x 2 )

13. Question f = 1 / T = 1 / 4.0ms = 1 / 0.004s = 250Hz (a) v = ± 2πf √(A 2 − x 2 ) maximum speed occurs when x = 0 v max = ± 2πf √(A 2 ) = ± 2πf A = ± 2π x 250 x 30μm = ± 2π x 250 x 0.000 030m maximum speed = 0.0471 ms -1 (b) v = ± 2πf √(A 2 − x 2 ) = ± 2π x 250 x √((0.000 030m) 2 − (0.000 015m) 2 ) = ± 500π x √((0.000 030m) 2 − (0.000 015m) 2 ) = ± 500π x √((9 x 10 -10 ) − (2.25 x 10 -10 )) = ± 500π x √(6.75 x 10 -10 ) = ± 500π x 2.598 x 10 -5 speed = 0.0408 ms -1 A body oscillating with SHM has a period of 4.0ms and amplitude of 30μm. Calculate (a) its maximum speed and (b) its speed when its displacement is 15μm.

14. Variation of x, v and a with time The acceleration, ax depends on the resultant force, Fspring on the mass. Note that the acceleration ax is always in the opposite direction to the displacement, X.

15. SHM time graphs T/4 T/2 3T/4 T 5T/4 3T/2 x = A cos (2π f t) a = - (2π f ) 2 A cos (2π f t) v = - 2π f A sin (2π f t) + A - A x v a time v max = ± 2π f A a max = ± (2π f ) 2 A Note: The velocity curve is the gradient of the displacement curve and the acceleration curve is the gradient of the velocity curve. a v x

16. SHM summary table DisplacementVelocityAcceleration + A0- (2π f ) 2 A 0± 2π f A0 - A0+ (2π f ) 2 A A ?B ? C ? E ? D? E ?

17. SHM graph question T/4 T/2 3T/4 T 5T/4 3T/2 a time a The graph below shows how the acceleration, a of an object undergoing SHM varies with time. Using the same time axis show how the displacement, x and velocity, v vary in time. x x acceleration is proportional to minus displacement v v velocity is the gradient of displacement

18. The spring-mass system If a mass, m is hung from a spring of spring constant, k and set into oscillation the time period, T of the oscillations is given by: T = 2π√(m / k) AS Reminder: Spring Constant, k: This is the force in newtons required to cause a change of length of one metre. k = F / ΔL unit of k = Nm -1 Mass on spring Mass on spring - Fendt

19. Question A spring extends by 6.0 cm when a mass of 4.0 kg is hung from it near the Earth’s surface (g = 9.8ms -2 ). If the mass is set into vertical oscillation state or calculate the period (a) near the Earth’s surface and (b) on the surface of the Moon where g = 1.7 ms -2. The spring constant, k is given by: k = F / ΔL = (4.0 x 9.8) / 0.060m = 653.3 Nm -1 (a) T = 2π√(m / k) = 2π √(4.0 / 653.3) = 2π √(0.006122) time period = 0.49s (b) g does not appear in the time period equation of the spring. Therefore the period on the Moon is the same = 0.49s

20. The simple pendulum A simple pendulum consists of: a point mass undergoing small oscillations (less than 10°) suspended from a fixed support by a massless, inextendable thread of length, L within a gravitational field of strength, g The time period, T is given by: T = 2π√(L / g) Simple PendulumSimple Pendulum - Fendt

21. Question Calculate: (a) the period of a pendulum of length 20cm on the Earth’s surface (g = 9.81ms -2 ) and (b) the pendulum length required to give a period of 1.00s on the surface of the Moon where g = 1.67ms -2. (a) T = 2π√(L / g) = 2π √(0.20m / 9.81ms -2 ) = 2π √(0.204) time period = 0.90s (b) T = 2π√(L / g) becomes: T 2 = 4π 2 (L / g) becomes: L = T 2 g / 4π 2 = ((1.00) 2 x 1.67) / 4π 2 length = 0.0423m (3.23cm)

22. Free oscillation A freely oscillating object oscillates with a constant amplitude. The total of the potential and kinetic energy of the object will remain constant. E P + E T = a constant This occurs when there are no frictional forces acting on the object such as air resistance.

23. Energy variation in free oscillation Position 1 – Maximum displacement Potential energy, E P = Total energy, E T Kinetic energy, E K = 0 2 1 equilibrium position 3 5 4 Position 2 E T = E P + E K Position 3 – Equilibrium position E P = 0 E K = Total energy, E T Position 4 E T = E P + E K Position 5 – Maximum displacement E P = Total energy, E T E K = 0 positionEPEP EKEK 1ETET 0 30ETET 5ETET 0

24. Question A simple pendulum consists of a mass of 50g attached to the end of a thread of length 60cm. Calculate: (a) the period of the pendulum (g = 9.81ms -2 ) and (b) the maximum height reached by the mass if the mass’s maximum speed is 1.2 ms -1. (a) T = 2π√(L / g) = 2π √(0.60m / 9.81ms -2 ) = 2π √(0.06116) time period = 1.56s (b) The maximum speed occurs at the equilibrium position, when the mass is at its lowest position so that its potential energy. E P = 0 kinetic energy, E K = ½ m v 2 = ½ x (0.050kg) x (1.2ms -1 ) 2 = 0.036 J The maximum height occurs when E P = E K so m x g x h = 0.036 J h = 0.036 / (0.050 x 9.81) maximum height = 0.074m

25. Energy variation with an oscillating spring The strain potential energy is given by: E P = ½ k x 2 Therefore the maximum potential energy of an oscillating spring system = ½ k A 2 = Total energy of the system, E T But: E T = E P + E K ½ k A 2 = ½ k x 2 + E K And so the kinetic energy is given by: E K = ½ k A 2 - ½ k v 2 E K = ½ k (A 2 - x 2 )

26. Energy versus displacement graphs The potential energy curve is parabolic, given by: E P = ½ k x 2 The kinetic energy curve is an inverted parabola, given by: E K = ½ k (A 2 - x 2 ) The total energy ‘curve’ is a horizontal line such that: E T = E P + E K - A 0 + A energy displacement, x EPEP EKEK ETET

27. Energy versus time graphs Displacement varies with time according to: x = A cos (2π f t) Therefore the potential energy curve is cosine squared, given by: E P = ½ k A 2 cos 2 (2π f t) ½ k A 2 = total energy, E T. and so: E P = E T cos 2 (2π f t) Kinetic energy is given by: EK = ET - EPEK = ET - EP E K = E T - E T cos 2 (2π f t) E K = E T (1 - cos 2 (2π f t)) E K = E T sin 2 (2π f t) E K = ½ k A 2 sin 2 (2π f t) ETET 0 T/4 T/2 3T/4 T energy time, t ½ kA 2 EPEP EKEK

28. Damping Damping occurs when frictional forces cause the amplitude of an oscillation to decrease. The amplitude falls to zero with the oscillating object finishing in its equilibrium position. The total of the potential and kinetic energy also decreases. The energy of the object is said to be dissipated as it is converted to thermal energy in the object and its surroundings.

29. Types of damping 1. Light Damping In this case the amplitude gradually decreases with time. The period of each oscillation will remain the same. The amplitude, A at time, t will be given by: A = A 0 exp (- C t) where A 0 = the initial amplitude and C = a constant depending on the system (eg air resistance)

30. 2. Critical Damping In this case the system returns to equilibrium, without overshooting, in the shortest possible time after it has been displaced from equilibrium. 3. Heavy Damping In this case the system returns to equilibrium more slowly than the critical damping case. light damping heavy damping critical damping displacement time A0A0

31. Forced oscillations All undamped systems of bodies have a frequency with which they oscillate if they are displaced from their equilibrium position. This frequency is called the natural frequency, f 0. Forced oscillation occurs when a system is made to oscillate by a periodic force. The system will oscillate with the applied frequency, f A of the periodic force. The amplitude of the driven system will depend on: 1. The damping of the system. 2. The difference between the applied and natural frequencies.

32. Resonance The maximum amplitude occurs when the applied frequency, f A is equal to the natural frequency, f 0 of the driven system. This is called resonance and the natural frequency is sometimes called the resonant frequency of the system.

33. Resonance curves applied force frequency, f A amplitude of driven system, A driving force amplitude f0f0 more damping light damping very light damping

34. Notes on the resonance curves If damping is increased then the amplitude of the driven system is decreased at all driving frequencies. If damping is decreased then the sharpness of the peak amplitude part of the curve increases. The amplitude of the driven system tends to be: - Equal to that of the driving system at very low frequencies. - Zero at very high frequencies. - Infinity (or the maximum possible) when f A is equal to f 0 as damping is reduced to zero.

35. Phase difference The driven system’s oscillations are always behind those of the driving system. The phase difference lag of the driven system depends on: 1. The damping of the system. 2. The difference between the applied and natural frequencies. At the resonant frequency the phase difference is π/2 (90°)

36. Phase difference curves applied force frequency, f A phase difference of driven system compared with driving system f0f0 more damping less damping - 90° - 180°

37. Examples of resonance Pushing a swing Musical instruments (eg stationary waves on strings) Tuned circuits in radios and TVs Orbital resonances of moons (eg Io and Europa around Jupiter) Wind driving overhead wires or bridges (Tacoma Narrows)

38. The Tacoma Narrows Bridge Collapse YouTube Videos: http://www.youtube.com/watch?v=3mclp9QmCGs – 4 minutes – with commentary http://www.youtube.com/watch?v=IqK2r5bPFTM&feature=related – 3 minutes – newsreel footage http://www.youtube.com/watch?v=j-zczJXSxnw&feature=related – 6 minutes - music background only An example of resonance caused by wind flow. Washington State USA, November 7th 1940.

39. Internet Links (SHM) Motion in 2DMotion in 2D - PhET - Learn about velocity and acceleration vectors. Move the ball with the mouse or let the simulation move the ball in four types of motion (2 types of linear, simple harmonic, circle). See the velocity and acceleration vectors change as the ball moves. Pendulum LabPendulum Lab - PhET - Play with one or two pendulums and discover how the period of a simple pendulum depends on the length of the string, the mass of the pendulum bob, and the amplitude of the swing. It's easy to measure the period using the photogate timer. You can vary friction and the strength of gravity. Use the pendulum to find the value of g on planet X. Notice the anharmonic behavior at large amplitude SHM & circular motion compared SHM & circular motion compared - NTNU SHM (spring) and circular motionSHM (spring) and circular motion - netfirms SHM with a spring & pendulumSHM with a spring & pendulum - Explore Science Mass on spring Mass on spring - Fendt Mass on a springMass on a spring - NTNU Mass on spring oscillations Mass on spring oscillations - NTNU Spring oscillation with varying mass and k Spring oscillation with varying mass and k - NTNU Spring oscillation showing graphs of s, v & a Spring oscillation showing graphs of s, v & a - NTNU Simple PendulumSimple Pendulum - Fendt Pendulum in an accelerated carPendulum in an accelerated car - NTNU Pendulum with rotating suspension pointPendulum with rotating suspension point - © 1998 Franz-Josef ElmerFranz-Josef Elmer

40. Internet Links (Forced Oscillation & Damping) Spring OscillationSpring Oscillation - PhET - A realistic mass and spring laboratory. Hang masses from springs and adjust the spring stiffness and damping. You can even slow time. Transport the lab to different planets. A chart shows the kinetic, potential, and thermal energy for each spring. Undamped and undriven pendulum Undamped and undriven pendulum - © 1998 Franz-Josef ElmerFranz-Josef Elmer Driven spring oscillator with one massDriven spring oscillator with one mass - Explore Science Driven spring oscillator with two massesDriven spring oscillator with two masses - Explore Science Forced oscillation Forced oscillation - Fendt Coupled PendualaCoupled Penduala - Fendt Pendulum driven by a periodic forcePendulum driven by a periodic force - © 1998 Franz-Josef ElmerFranz-Josef Elmer Horizontally driven pendulumHorizontally driven pendulum - © 1998 Franz-Josef ElmerFranz-Josef Elmer Vertically driven pendulumVertically driven pendulum - © 1998 Franz-Josef ElmerFranz-Josef Elmer Resonance in a stringResonance in a string - netfirms RCL Resonance RCL Resonance - netfirms

41. Core Notes from Breithaupt pages 34 to 49 1.What is an oscillation? 2.In the context of oscillations explain the meaning of the following: (a) equilibrium; (b) displacement; (c) amplitude; (d) time period; (e) frequency; (f) angular frequency and (g) phase difference. 3.Explain the conditions (in terms of acceleration and displacement) required for an object to be undergoing simple harmonic motion. 4.State the defining equation for SHM shown on page 37. 5.Sketch the graphs shown on page 36 and explain the mathematical relationships between them. 6.(a) State the equation for the time period for a mass-spring system. (b) Calculate the mass required to give a mass-spring system of spring constant 25N/m a time period of one second. 7.(a) State the equation for the time period for a simple pendulum. (b) Calculate the length required to give a pendulum a time period of one second near the Earth’s surface. 8.How, if at all would your answers to Q6(b) and Q7(b) be different if you were on the Moon’s surface? (g Moon = 1.7 N/kg) 9.Explain how potential, kinetic and total energy vary during the oscillation of a simple pendulum. (assume no resistive forces). 10.What is damping? How would it change your answers to Q9? 11.What is meant by ‘resonance’? 12.Sketch and explain the resonance curves on page 47. 13.Explain how damping affects the phase difference between the driving force and the driven system.

42. Notes from Breithaupt pages 34 & 35 Oscillations 1.What is an oscillation? 2.In the context of oscillations explain the meaning of the following: (a) equilibrium; (b) displacement; (c) amplitude; (d) time period; (e) frequency; (f) angular frequency and (g) phase difference. 3.Try the summary questions on page 35

43. Notes from Breithaupt pages 36 & 37 The principles of simple harmonic motion 1.Explain the conditions (in terms of acceleration and displacement) required for an object to be undergoing simple harmonic motion. 2.State the defining equation for SHM shown on page 37. 3.Sketch the graphs shown on page 36 and explain the mathematical relationships between them. 4.Try the summary questions on page 37

44. Notes from Breithaupt pages 38 & 39 More about sine waves 1.Explain the ways in which circular motion and SHM correspond with each other. 2.Try the summary questions on page 39

45. Notes from Breithaupt pages 40 to 43 Applications of simple harmonic motion 1.(a) State the equation for the time period for a mass-spring system. (b) Calculate the mass required to give a mass-spring system of spring constant 25N/m a time period of one second. 2.(a) State the equation for the time period for a simple pendulum. (b) Calculate the length required to give a pendulum a time period of one second near the Earth’s surface. 3.How, if at all would your answers to Q1(b) and Q2(b) be different if you were on the Moon’s surface? (g Moon = 1.7 N/kg) 4.Derive the equations for (a) the spring-mass system and (b) the simple pendulum. 5.Try the summary questions on page 43

46. Notes from Breithaupt pages 44 to 46 Energy and simple harmonic motion 1.Explain how potential, kinetic and total energy vary during the oscillation of a simple pendulum. (assume no resistive forces). 2.What is damping? How would it change your answers to Q1? 3.Repeat Q1, this time for a mass-spring system. 4.Describe with the aid of suitable graphs the different types of damping. 5.Try the summary questions on page 46

47. Notes from Breithaupt pages 47 to 49 Forced oscillations and resonance 1.What is meant by ‘resonance’? 2.Sketch and explain the resonance curves on page 47. 3.Explain how damping affects the phase difference between the driving force and the driven system. 4.Explain the demonstration ‘Barton’s Pendulum’ on page 48. 5.Give and explain an example of resonance with bridges. 6.Try the summary questions on page 49