Chapter 17 Electric Field

Chapter 14 OscillationsA2Mr. Chong Kwai KunLearning OutcomesCandidates should be able to:(a) describe simple examples of free oscillations(b) investigate the motion of an oscillator using experimental and graphical methods(c) understand and use the terms amplitude, period, frequency, angular frequency and phase difference and express the period in terms of both frequency and angular frequency(d) recognise and use the equation a = 2x as the defining equation of simple harmonic motion(e) recall and use x = x0sint as a solution to the equation a = 2x(f) recognise and use v = v0cos t, v = ( )(g) describe, with graphical illustrations, the changes in displacement, velocity and acceleration during simple harmonic motion

Learning OutcomesCandidates should be able to:(h) describe the interchange between kinetic and potential energy during simple harmonic motion(i) describe practical examples of damped oscillations with particular reference to the effects of the degree of damping and the importance of critical damping in cases such as a car suspension system(j) describe practical examples of forced oscillations and resonance(k) describe graphically how the amplitude of a forced oscillation changes with frequency near to the natural frequency of the system, and understand qualitatively the factors that determine the frequency response and sharpness of the resonance(l) show an appreciation that there are some circumstances in which resonance is useful and other circumstances in which resonance should be avoided.Chapter 14.1Simple HarmonicMotionOscillatory MotionAny motion that repeats itself after equal intervals of time is called periodic motion. If an object in periodic motion moves back and forth over the same path, the motion is called oscillatory or vibratory motionFree and Forced VibrationsUses of OscillationsPleasureMusicKeeping timeOtherFree OscillationsForced Vibrations

Observing oscillationMass-spring systemA pendulumA loudspeaker cone

8Simple harmonic motion S.H.M.A special kind of oscillation

XYOX, Y : extreme pointsO: centre of oscillation / equilibrium positionA: amplitudeAA special relation between the displacement and acceleration of the particleAn oscillation is a repeated motion about a fixed point which known as the equilibrium position.The amplitude, A of an oscillation is equal to the maximum displacement, x

Frequency, f in hertz is equal to the number of complete oscillations per second.

also: f = 1 / T

Angular frequency, in radians per second is given by:

= 2 for = 2 / T

equilibrium positionx = 0x = + Ax = 0x = - Ax = 09Simple Harmonic MotionA 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.TensionGravityResultant (restoring force)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.MTgRestoring forceSimple Harmonic MotionSimple Harmonic Motion

When the force acting on the particle is directly proportional to the displacement and opposite in direction, the motion is said to be linear simple harmonic motion

The pattern of SHM motionThe pattern of SHM is the same as the side view of an object moving at a constant speed around a circular pathThe 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- A13Conditions required for SHMWhen 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

14Acceleration variation of SHMThe constant k is equal to (2f )2 or 2 or (2/T)2

Therefore:a = - (2f )2 x (given on data sheet)ora = - 2 xora = - (2/T)2 xxa+ A- A- 2 x+ 2 xgradient = - 2 or - (2f )2 15Question 1A 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 = - (2f )2 xmaximum acceleration is when x = A (the amplitude)a = - (2f )2 A = - (2 x 0.6667Hz)2 x 0.015mmaximum acceleration = 0.88 ms-2

16Simple Harmonic MotionA mass on a spring has a displacement as a function of time that is a sine or cosine curve:

Here, A is called the amplitude of the motion.SHM DisplacementIf we call the period of the motion T this is the time to complete one full cycle we can write the position as a function of time:

It is then straightforward to show that the position at time t + T is the same as the position at time t, as we would expect.

Uniform Circular Motion and SHMAn object in simple harmonic motion has the same motion as one component of an object in uniform circular motion:The object in circular motion has an angular speed of

where T is the period of motion of the object in simple harmonic motion.Uniform Circular Motion and SHMThe position as a function of time:

The angular frequency:

The velocity as a function of time:

And the acceleration:

Both of these are found by taking components of the circular motion quantities.Velocity and Acceleration of SHM21Velocity equations of SHMThe velocity, v of an object oscillating with SHM varies with displacement, x according to the equations:

v = 2f (A2 x 2) orv = (A2 x 2)orv = (2 / T) (A2 x 2)22Question 2A 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 (2 f t) = 4.0mm x cos (2 50 x 2.0ms) = 4.0mm x cos (2 50 x 0.0020s) = 0.6283 displacement = 0.628 mm

(b) a = - (2f )2 x = - (2 x 50)2 x 0.6283mm = - (100)2 x 0.0006283macceleration = - 62 ms-223Question 3f = 1 / T= 1 / 4.0ms= 1 / 0.004s= 250Hz

(a) v = 2f (A2 x 2) maximum speed occurs when x = 0vmax = 2f (A2) = 2f A= 2 x 250 x 30m= 2 x 250 x 0.000 030mmaximum speed = 0.0471 ms-1(b) v = 2f (A2 x 2) = 2 x 250 x ((0.000 030m)2 (0.000 015m)2) = 500 x ((0.000 030m)2 (0.000 015m)2) = 500 x ((9 x 10 -10) (2.25 x 10 -10))= 500 x (6.75 x 10 -10)= 500 x 2.598 x 10 -5speed = 0.0408 ms-1

A body oscillating with SHM has a period of 4.0ms and amplitude of 30m. Calculate (a) its maximum speed and (b) its speed when its displacement is 15m.24The spring-mass systemIf a mass, m is hung from a spring of spring constant, k and set into oscillation. Since the force on a mass on a spring is proportional to the displacement,

The Period of a Mass on a SpringTherefore, the period is

Example A particle moving with S.H.M. has velocities of 4 cm s-1 and 3 cm s-1 at distances of 3 cm and 4 cm respectively from its equilibrium. Find(a) the amplitude of the oscillation

Solution:By v2 = w2(A2 x2)when x = 3 cm, v = 4 cm s-1,x = 4 cm, v = 3 cm s-1. 42 = w2(A2 32) --- (1) 32 = w2(A2 42) --- (2)(1)/(2):16/9 = (A2 9) / (A2 16)9A2 81 = 16 A2 - 256 A2 = 25A = 5 cm amplitude = 5 cm274 ms-13 ms-1O(b)the period,(c)the velocity of the particle as it passes through the equilibrium position.(b)Put A = 5 cm into (1) 42 = w2(52 32) w2 = 1 w = 1 rad s-1T = 2p/w = 2p s(c)at equilibrium position, x = 0By v2 = w2(A2 x2)v2 = 12(52 02)v = 5 cm s-128Energy Conservation in Oscillatory MotionIn an ideal system with no nonconservative forces, the total mechanical energy is conserved. For a mass on a spring:

Since we know the position and velocity as functions of time, we can find the maximum kinetic and potential energies:

Energy Conservation in Oscillatory MotionAs a function of time,

So the total energy is constant; as the kinetic energy increases, the potential energy decreases, and vice versa.Energy Conservation in Oscillatory Motion

This diagram shows how the energy transforms from potential to kinetic and back, while the total energy remains the same.Energy variation with an oscillating springThe strain potential energy is given by:EP = k x2Therefore the maximum potential energy of an oscillating spring system= k A2= Total energy of the system, ET

But: ET = EP + EK k A2 = k x2 + EKAnd so the kinetic energy is given by:EK = k A2 - k v2

EK = k (A2 - x2)32Energy versus displacement graphsThe potential energy curve is parabolic, given by:EP = k x2

The kinetic energy curve is an inverted parabola, given by:EK = k (A2 - x2)

The total energy curve is a horizontal line such that: ET = EP + EK

- A 0 + Aenergydisplacement, xEPEKET33Energy versus time graphsDisplacement varies with time according to:x = A cos (2 f t)Therefore the potential energy curve is cosine squared, given by:EP = k A2 cos2 (2 f t)

k A2 = total energy, ET.and so: EP = ET cos2 (2 f t)Kinetic energy is given by:EK = ET - EPEK = ET - ET cos2 (2 f t)EK = ET (1 - cos2 (2 f t))EK = ET sin2 (2 f t)EK = k A2 sin2 (2 f t)

ET0 T/4 T/2 3T/4 Tenergytime, t kA2EPEK34

The PendulumA simple pendulum consists of a mass m (of negligible size) suspended by a string or rod of length L (and negligible mass).The angle it makes with the vertical varies with time as a sine or cosine.The PendulumLooking at the forces on the pendulum bob, we see that the restoring force is proportional to sin , whereas the restoring force for a spring is proportional to the displacement (which is in this case).

The PendulumHowever, for small angles, sin and are approximately equal.

The PendulumSubstituting for sin allows us to treat the pendulum in a mathematically identical way to the mass on a spring. Therefore, we find that the period of a pendulum depends only on the length of the string:

39A simple pendulum has a period of 2 s and an amplitude of swing 5 cm. Calculate the maximum magnitudes of (a)velocity, and (b)acceleration of the bob.Solution:(a)maximum magnitude of velocity = wA = p(5) = 5p cm s-1 (b)maximum magnitude of velocity = w2A = 5p2 cm s-2

Question 4Question 4Calculate: the period of a pendulum of length 20cm on the Earths 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 = 2(L / g) = 2 (0.20m / 9.81ms-2) = 2 (0.204) time period = 0.90s

(b) T = 2(L / g) T2 = 42 (L / g) L = T2g / 42 = ((1.00)2 x 1.67) / 42 length = 0.0423m (3.23cm)40An mass oscillates with an amplitude of 4.00 m, a frequency of 0.5 Hz and a phase angle of p/4. What is the period T?Write an equation for the displacement of the particle.Calculate the velocity and acceleration of the object at any time t.Question 5

(d) Determine the position, velocity and acceleration of the object at time t = 1.00s. (e) Calculate the maximum velocity and acceleration of the object.41A spring stretches by 3.90 cm when a 10.0 g mass is hung from it. A 25.0 g mass attached to this spring oscillates in simple harmonic motion.

Calculate the period of the motion.Calculate frequency and the angular velocity of the motion.Question 642Question 7A simple pendulum consists of a mass of 50g attached to the end of a thread of length 60cm. Calculate: the period of the pendulum (g = 9.81ms-2) the maximum height reached by the mass if the masss maximum speed is 1.2 ms-1.

(a) T = 2(L / g) = 2 (0.60m / 9.81ms-2) = 2 (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. EP = 0

kinetic energy, EK = m v2= x (0.050) x (1.2)2= 0.036 J

The maximum height occurs when EP = EK

so m x g x h = 0.036 Jh = 0.036 / (0.050 x 9.81)maximum height = 0.074m43Free oscillationA freely oscillating object oscillates with a constant amplitude.

The total of the potential and kinetic energy of the object will remain constant.EP + ET = a constant

This occurs when there are no frictional forces acting on the object such as air resistance.

44DampingDamping 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.

45Damped Oscillations

This exponential decrease is shown in the figure:

Damped OscillationsThe previous image shows a system that is underdamped it goes through multiple oscillations before coming to rest. A critically damped system is one that relaxes back to the equilibrium position without oscillating and in minimum time; an overdamped system will also not oscillate but is damped so heavily that it takes longer to reach equilibrium.Types of damping1. Light DampingIn 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 = A0 exp (- c t)where A0 = the initial amplitude and c = a constant depending on the system (eg air resistance)

482. Critical DampingIn this case the system returns to equilibrium, without overshooting, in the shortest possible time after it has been displaced from equilibrium.

3. Heavy DampingIn this case the system returns to equilibrium more slowly than the critical damping case.light dampingheavy dampingcritical dampingdisplacementtimeA049Forced oscillationsAll 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, f0.

Forced oscillation occurs when a system is made to oscillate by a periodic force. The system will oscillate with the applied frequency, fA 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.50Driven Oscillations and ResonanceAn oscillation can be driven by an oscillating driving force; the frequency of the driving force may or may not be the same as the natural frequency of the system.

ResonanceThe maximum amplitude occurs when the applied frequency, fA is equal to the natural frequency, f0 of the driven system.

This is called resonance and the natural frequency is sometimes called the resonant frequency of the system.52Resonance curvesapplied force frequency, fAamplitude of driven system, Adriving force amplitudef0more dampinglight dampingvery light damping53Notes on the resonance curvesIf 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 the driving system at very low frequencies.- Zero at very high frequencies.- Infinity (or the maximum possible) when fA is equal to f0 as damping is reduced to zero.54The Tacoma Narrows Bridge CollapseYouTube Videos:http://www.youtube.com/watch?v=3mclp9QmCGs 4 minutes with commentaryhttp://www.youtube.com/watch?v=IqK2r5bPFTM&feature=related 3 minutes newsreel footagehttp://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 1940.55Summary The amplitude is the maximum displacement from equilibrium. Position as a function of time:

Velocity as a function of time:

Summary Acceleration as a function of time:

Period of a mass on a spring:

Total energy in simple harmonic motion:

Summary Potential energy as a function of time:

Kinetic energy as a function of time:

A simple pendulum with small amplitude exhibits simple harmonic motionSummary Period of a simple pendulum:

Summary Oscillations where there is a nonconservative force are called damped. Underdamped: the amplitude decreases exponentially with time:

Critically damped: no oscillations; system relaxes back to equilibrium in minimum time Overdamped: also no oscillations, but slower than critical dampingSummary An oscillating system may be driven by an external force This force may replace energy lost to friction, or may cause the amplitude to increase greatly at resonance Resonance occurs when the driving frequency is equal to the natural frequency of the system