PH10005/53: VIBRATIONS, WAVES & OPTICS

Week Starts Monday9:15

3WN 2.1

Thursday12:15

CB 1.11

Friday9:15

5W 2.319 3 Feb Lecture 1 Lecture 220 10 Feb Lecture 3 Lecture 421 17 Feb Lecture 5 Lecture 6 Prob class 122 24 Feb Lecture 7 Lecture 823 3 Mar Lecture 9 Lecture 10 Prob class 224 10 Mar Lecture 11 Lecture 12 Prob class 325 17 Mar Lecture 13 Lecture 14 Prob class 426 24 Mar Lecture 15 Lecture 16 Mock Exam27 31 Mar Lecture 17 Lecture 18 Prob class 528 7 Apr Lecture 19 Lecture 2029 14 Apr Lecture 21 Prob class 6

& revisionHolliday

Lecture’s contact details: Sergey Gordeev: Office 3W4.17 E-mail: s.gordeev@bath.ac.uk.

Recommended textbooks1. P.A. Tipler& G.P. Mosca. Physics for Scientists and Engineers

(W.H. Freeman and Company, 4th or later edition).This textbook contains most of the basic syllabus, but you will need to supplement them with additional material.

2. H.J. Pain. The Physics of Vibrations and Waves (Wiley, Chic hetser, 6th or any other edition), 2005.

This book covers all the required material but in a more mathematical way than Tipler.

3. E. Hecht. Optics (Addison Wesley, any edition)Very good book on optics.

4. D. Halliday, R. Resnick & J. Walker. Fundamentals of Physics (Wiley, 8th

edition, extended), 2008Similar to Tipler & Mosca, contains most of the basic syllabus.

COURSE STRUCTURE

I. Oscillations II. Mechanical WavesIII. Geometrical OpticsIV. Wave Optics

_______________________________________________________

1

I. OSCILLATION I. Simple Harmonic Motion

§ Suggested ReadingPain Chapter 1 Simple Harmonic OscillationsTipler Chapter 14 Oscillations

1.1. IntroductionPeriodic motion is motion of an object that regularly repeats, i.e. the object returns to a given position after a fixed time interval.

Examples:Earth; Moon, molecules in a solid; Electromagnetic waves (light); clock; loudspeakers; bridges; tall buildings.

Very important kind of periodic motion is a Simple Harmonic Motion (SHM).

1.2. Mechanics of Simple Harmonic MotionIn oscillating mechanical systems there is always a force that acts to reduce displacement of an object and return the system to equilibrium. This force is called the restoring force, Fr .

A typical system that exhibits SHM:

In equilibrium - no force on the object.

2

SHM occurs whenever the restoring force is linearly proportional to the displacement from equilibrium and opposite to the direction of the displacement, i.e. Fr-x.

Model:

a) no friction;

b) mass of spring is negligibly small in comparison with the mass of block

When the object is displaced, the spring exerts a restoring force -kx, as given by Hooke’s Law:

(1.1)

where k is the constant of proportionality between the displacement and the restoring force, called the spring constant (or force constant).

The minus sign in Hooke’s law arises because the force is opposite the direction of the displacement.

Using Newton’s Second Law of Motion we get

(1.2)

This can be written as:

(1.3)

If we calculate the dimensions of k/m then we find that this has a dimension of (time)-2, so

(1.4)

where is an angular frequency.

Eq. 1.4 is a second order differential equation. It is often referred to as the differential equation of Simple Harmonic Motion.

Its solution is(1.5)

where A and δ are defined by the initial conditions (i.e. displacement and velocity at t=0).

According Eq. 1.5, function x(t) has the form:

Units: Period: T in ‘seconds’, Frequency: f=1/T in ‘Hertz’, Angular frequency: in ‘radians per second’.

3

1.3 Displacement, Velocity and AccelerationIf Then velocity is (1.6)and acceleration is

(1.7)

Graphs (for δ=0):

Thus,Velocity is π/2 out of phase with the displacement.

Velocity is a maximum when displacement is zero.Velocity is zero when displacement is a maximum/

Acceleration is proportional to displacement and acts in the opposite sense to the displacement.

Note that the minimum isn’t the zero point. The displacement, velocity and acceleration oscillate about zero with maximum for positive values and a minimum at their largest negative values.

The frequency and period are related to the stiffness k of the spring

and the particle mass: and .

4

Note that spring constant k is only a constant for small displacements .

Large displacements would cause permanent deformation of the spring and the system would not oscillate with SHM. This reminds us that SHM is only found for small displacements.

ExampleAn air-track glider attached to a spring oscillates with a period of 1.5s. At t=0 the glider is 5cm left of the equilibrium position and moving to the right at 36.3 cm/s. (a) What are the amplitude and initial phase of the oscillations?(b) Write down an expression that describes position of the glider as a function of time. (c) What is the glider’s position at t=0.5s?

5

1.4. Further examples of Simple Harmonic MotionFrom our basic definition, we observe SHM when the restoring force is linearly proportional to displacement. Then, for any simple harmonic motion, we can write out an expression for the force constant, k

so that (1.9)

1.4.a. An Object on a Vertical SpringShould take into account gravitational force mg in addition to the force of the spring Fy=-ky, where y is the vertical displacement. In equilibrium, the spring is extended by a length y0.

6

Apply Newton’s second law to the equilibrium situation to find yo:

(1.10)For oscillations:

(1.11)

This differs from Eq. 1.2 by the constant term mg. We can handle this extra term by changing to a new variable

y=y-y0.

For y, Eq. 1.11 reduces to

(1.12)

and its solution is Eq. 1.5 with x replaced by y:(1.5b)

1.4b. The Simple Pendulum

Model: m is a point mass connected by a massless, inextensible string.

The restoring force along the arc of the circle:(1.13)

7

The tangential component of the acceleration: .

The tangential component of Newton’s second law is thus

(1.14)

For small angles :

(1.15)

and with .

Thus for small displacements, the period of the pendulum is independent of mass of the bob and depends only on the length of the pendulum.

Notice that the pendulum does not exhibit true SHM for any angle. If the angle is less than 10º, the motion is close to and can be modelled as simple harmonic.

1.5. Energy in Simple Harmonic MotionWe know that for any conservative force that there is a direct link to the potential energy of the system, given by

(1.16)As for SHM , we can write that and

(1.17)

Parabolic dependence of potential energy on displacement is a general result for anything moving with SHM.

The kinetic energy of the oscillator (1.18)

and the total energy at any moment of time is(1.19)

For any oscillator, we can write

(1.20)

Therefore

8

(1.21)

From this we can conclude that energy in SHM is conserved.

1.6. The General Applicability of Simple Harmonic Motion In SHM, the restoring force is zero at the equilibrium point. As

the potential energy at this point has its minimum.

For small displacements, we may expand U(x) in a Maclaurin-Taylor series:

(1.22)

If we choose the zero point of potential energy such that U(0)=0 then two first terms in the equation above are equal to zero. Thus we can write that, approximately,

(1.23)

where .

The small-amplitude motion near almost any point of equilibrium is described approximately by a parabolic potential well. With increasing amplitude of oscillations, contributions of terms with higher derivates in Eq 1.22 becomes larger and larger and the motion will be anharmonic.

Example A 0.5-kg cart connected to alight spring for which the force constant is 20N/m oscillates on a horizontal, frictionless airtrack.

a) Calculate the total energy of the system and the maximum speed of the cart if the amplitude of the motion is 3 cm.

(b) What is the velocity of the cart when the position is 2 cm?(c) Compute the kinetic and potential energies of the system when the position is 2 cm.

9

SUMMARY

Simple Harmonic MotionSimple harmonic motion will occur whenever an object moves from equilibrium under a restoring force that is linearly proportional to the displacement from equilibrium, i.e. it moves with a linear restoring force.

Defining Equation for SHM

General Solution

Proving that a motion is simple harmonic motion

We can demonstrate that a motion is SHM if we can show that, following the definition above, Fx=-kx where k is a constant - the force constant of the motion.

Energies in Simple Harmonic MotionPotential energy:

Kinetic energy:

Total Energy:

Universality of Simple Harmonic MotionFor small displacements from equilibrium, in most potential wells, the motion of a displaced mass can be approximated to be simple harmonic motion with a force constant, k=U(0).

10

II. Damped Harmonic Oscillations§ Suggested Reading

Pain Chapter 2 Damped Simple Harmonic MotionTipler Chapter 14 Oscillations

IntroductionIn real oscillating systems the mechanical energy of the system diminishes in time, and the motion is said to be damped.

The mechanism of damping is viscous friction. We will assume that the magnitude of the damping force can be taken to depend on the relative speed v between the parts of the system:

(2.1)

where b is the damping force constant. This is normally a good approximation for low velocities.

Example:

11

Fig. 2.1 A damped oscillator. The motion is damped by the plunger immersed in the liquid.

2.1. Differential Equation for Damped Harmonic motion

Applying Newton’s second law we get:

(2.2)where is the restoring force and is the damping force.

If the oscillations occur along the x-axis only the above equations reduces to

(2.3)from which

(2.4)

The latter is called the equation of motion of a damped harmonic oscillator.

This equation is often presented in the form

(2.5)

which simplifies the subsequent algebra. Here is the damping constant and is the circular frequency of the oscillations which would have occurred had there been no damping.

12

2.2. Solutions for Damped Oscillations

Eq.2.5 is a second order differential equation. Its general solution, derived fully in the maths module.

A reminder of the technique (you must be able to work through this on your own!):

For the case of the trial solution (2.6)

gives auxiliary equation (2.7)

So has two values (2.8)

There are 3 cases to be considered:(a) (heavy damping or overdamping)(b) (critical damping)(c) (light damping)

13

(a) Overdamping ( )The condition means that (show this!) which implies that the viscous friction force term dominates the restoring force causing heavy damping of the oscillator.

Eq.2.8 gives two real distinct roots 1 and 2.

According to Eq.2.6 the general solution is (2.9)

where A1 and A2 are arbitrary constants of integration which have to be determined from the initial conditions, such as the position and velocity of the particle at some instant.

Fig. 2.2. Heavy damping. Mass displaced 9 mm and released from rest: 0=0.2 s-1, =0.5 s-1, A1=9.4 mm and A2=-0.4 mm.

Note there are no oscillations. The displaced mass creeps back, or relaxes, to its equilibrium position very slowly.

14

(b) Critical damping ( )This solution leads to a simplification of Eq. 2.5 such that

(2.10)

Again, A and B are arbitrary constants determined by the set of initial conditions. The exponential term is never zero. The term (At+B), for some initial conditions, can give one pass of the oscillator through its equilibrium position but the motion is not recognisable as an oscillationA typical motion of this kind is shown in the Fig. 2.3 below.

This motion results in a fast decay of the motion back to its equilibrium position.

Fig. 2.3. Critical damping. The return to the equilibrium position is faster than in the case of overdamping.

Critical damping is employed in many practical circumstances, in shock absorbers in suspension systems, in the design of electrical meters etc.

15

(c) Weak damping ( )This is physically the most interesting case. The solution (2.9) is still valid, but the exponents are complex quantities:

(2.11)

For generality, A1 and A2 must be complex quantities, but clearly (A1+A2) and i(A1-A2) must be real. Putting these respectively as D and C, we have

(2.12)

which is one way of expressing the general solution.

An alternative form of the solution is obtained by introducing the phase constant ,

(2.13)

where a is the amplitude of the motion at time t=0.

This solution is conveniently expressed as(2.14)

where(2.15)

16

Now this is similar to Eq. 1.5, for free oscillations, but the amplitude term A has been replaced by , and the circular frequency term has become .

The amplitude of the motion decreases exponentially with time. Also the frequency of the oscillations is smaller (<0).

The period of this motion can be defined as the time between the points U and W in the figure above:

(2.16)

17

2.3. Logarithmic Decrement

The rate at which the amplitude of oscillations dies away is characterised by logarithmic decrement of the decay.

It is defined as the natural logarithm of the ratio of two successive maximum displacements (i.e. maxima that are separated by one period):

(2.17)

For week damping, according to Eq. 2.14, maximum displacements corresponding to times and approximately correspond to the points where :

(2.18)From which

(2.19)

18

2.4. The Quality Factor or Q-Factor

The quality of oscillator is determined by the fraction of its energy that is lost per cycle of its displacement.

The quality factor is defined as the inverse of the fractional loss of energy:

(2.20)

where E is the total energy at time t and is the energy loss over one period (at time t).

Q-factor is a dimensionless quantity (a number). Very lightly damped oscillators have a Q>>1. For examples, a tuning fork might have a Q-factor in the range of 103-104.

Let us estimate the energy of oscillator when it reaches its maximum deviations at time t and t+T. As v(t)= v(t+T)=0 the energy of oscillator is equal to the elastic

potential energy stored in the spring .

Taking into account that and that at points of maximum deviation we get

(2.21)

Therefore, the total energy of the system decays exponentially with time.

19

The rate of change of energy is

(2.22)

The energy loss per one period is(2.23)

Therefore, the Q-factor can be written as

(2.24)

Notice that Q is independent of time and the actual energy possessed by the vibrating system at any given instant.

2.5. Decay timeThe differential equation 2.22 for energy has the following solution

(2.25)where E0 is the initial energy of the system.

Eq. 2.25 can be also written in the form (2.26)

where =1/(2) is the decay constant.

How many periods exist in the decay time, ?

That is easy to calculate:

(2.27)

So we can see that the number of periods in the time constant of the decay is ~Q/6.Example When middle C on a piano (frequency 262 Hz) is struck, it loses half its energy after 4s.

(a) What is the decay constant, τ ?(b) What is the Q-factor for this oscillation?(a) What is the fractional loss of energy per cycle?

20

SUMMARY

We are assuming a damping force that is proportional to the velocity of oscillator body.

This gives an exponential decay in the energy of the oscillator, where coefficient is called the damping constant.

Logarithmic decrement =

where an and an+1 are two successive amplitudes one period apart.

The Q-factor is defined by the fractional energy loss per cycle and found to be

The equation of motion for a damped harmonic oscillator is

The solution for an underdamped oscillator is

where , which is a lower frequency than an undamped oscillator with the same force constant.

21

III. Forced Vibrations and Resonance

§ Suggested ReadingPain Chapter The Forced OscillatorTipler Chapter 14 Oscillations

3.1 IntroductionOscillating systems have natural frequencies ( ) at which they vibrate when excited. Examples: bells, drums, pendulums etc.

If we want to make something vibrate at a different frequency (Ω) then we must apply a periodic driving force acting with the required frequency Ω.

If the frequency of the driving force is equal (or very close) to that of the natural frequency ω0 then energy is efficiently transferred from the driving force to the oscillator increasing the amplutude . This phenomenon is called resonance.

Examples of resonance: A child’s swing An opera singer who causes a glass to crack in response to

the driving force of the sound waves from her voice. Tacoma Narrows bridge (Tipler p. 503)

3.2 Differential Equation for Forced Oscillations

Restoring force: -kxDamping force: Driving force: F0 sin(Ωt)

Appying Newton’s second law:

(3.1)

This gives the equation describing a harmonically driven oscillator with linear damping:

(3.2)

22

0 x

Fr Fd

F0 sinΩt

Eq. 3.2 will be easier to handle if we make some substitutions. Similar to the previous chapter: -

Damping constant: Natural angular frequency: ,

Maximum driving acceleration:

This gives the Differential Equation of Forced Oscillations (make sure that you can show this)

(3.3)

We are going to find the steady state response of the system ignoring the initial transient response. In fact, the transient motion is a combination of solutions to the equation for the undriven oscillator (found in the previous lecture) and the steady state solution that we going to obtain.

3.3. The Trial SolutionWe will assume a steady state motion of the form

(3.4)

The response of the system to the driving will have an amplitude, A, and a phase lag, , between the steady state motion and the driving force.

3.4. Steady State Response of Driven Oscillator[Start of proof] (the proof is examinable ! )The final solution must satisfy,

(3.3a)

If then

(3.5)

Substituting these values into Eq. 3.3a, we find

23

(3.6)

To solve this equation and find expressions for A and we will compare the coefficients of the and on the left hand and right hand side of the expression above. If the equation is correct for all frequencies and times then the coefficients must always be equal.

First, we must re-express the driving term in terms of the response.

(3.7)

Using trigonometric identity

we get

Substituting this into Eq. 3.6, we find

(3.8)

This is an identity valid for all times, t, so we can equate the coefficients. We find that,

(3.9a)

(3.9b)

Therefore:

(3.9c)

and sin2+cos2=1 so we can get:

Amplitude and Phase Response of Driven Oscillator:

AMPLITUDE of RESPONSE

24

(3.10)

From we get

PHASE LAG of RESPONSE

(3.11)

[END OF PROOF]

We will now investigate the form of the frequency response at low frequencies, high frequencies and at the oscillator’s natural frequency.

25

Special cases(a) Low Frequency Limit: <<0

Amplitude: (3.12a)

Phase lag: =0 (3.12b)

Solution: (3.12c)

The driven body moves in phase with the driving force with an amplitude that is inversely proportional to the square of its natural frequency.

(b) Resonance Region: 0

(strictly speaking, the resonance frequency is )

Amplitude: (3.13a)

Phase Lag: (3.13b)

Solution: (3.13c)

(c) High Frequency Limit: >>0

Amplitude: (3.14a)

Phase Lag: (3.14b)

Solution: (3.14c)

26

3.6. Details of the Resonant Response

We can see that there will be a large amplitude when a system with a small damping constant is driven at its natural frequency. The frequency with the greatest system response is defined as the resonance frequency for the system.

When the damping is small (large Q), the width of the resonance curve is narrow. If the damping is greater then the oscillator absorbs energy over a wider range of frequencies and thus the width is much greater. For small values of damping the relation for the Quality factor, Q is:

(3.15)Thus the Q factor is a direct measure of the sharpness of resonance.

How much extra motion do we get at resonance? This is easy to calculate.

(3.16)

§ Summary of Important Frequencies So, this now gives us three important frequencies to remember for oscillating systems.

27

Simple Harmonic

Motion

Damped Harmonic Motion Driven Harmonic Oscillator

Natural frequency

Underdamped oscillator frequency

Resonant frequency

Example 1Find an expression for the resonance frequency of a driven oscillator in terms of the parameters of the defining differential equation

which gives a solution for the driven amplitude,

.

Example 2An oscillator consists of a mass of 0.02kg suspended from a spring with a spring constant of 50 N/m. It is subjected to a damping force of the form where v is the velocity and b=0.5 kg s-1, and is driven by a harmonically varying force of amplitude 4N and frequency 550 rpm. Calculate the amplitude and phase lag of the resulting steady state vibration.

28

Summary The differential equation for a driven damped harmonic oscillator is: -

The STEADY STATE response to the driving can be found using the trial solution

which gives (you must be able to prove this)

AMPLITUDE of RESPONSE

PHASE LAG of RESPONSE

This enables us to define the Q-factor for a driven oscillator by:

The idea of resonance is useful for many branches of physics and scientists often want to obtain resonance in order to detect something or avoid resonance to stop something from breaking.

PROBLEM SHEET 1

Simple Harmonic Motion (SHM)

1. (Basic) A boat at anchor bobs up and down with the waves. The boats moves 5 cm above and 5 cm below its equilibrium position, and makes one complete up-and-down cycle every 4 s. What are the amplitude, period, frequency, and angular frequency of the motion?

29

[Ans: 5 cm, 4 s, 0.25 Hz, 1.57 rad/s]

2. (Standard) An object oscillates with a simple harmonic motion along the x axis. Its position varies with time according to the equation

where all parameters are in SI units (meters, radians, seconds). (a) Determine the amplitude, frequency, and period of the motion.(b) Calculate the velocity and acceleration of the object at any time t.(c) Determine the position, velocity, and acceleration of the object at t=1 s.(d) Determine the maximum speed and maximum acceleration.(f) Determine the displacement of the objects between t=0 and t=1 s.

[Ans: (b) , (c) -2.83 m, 8.89 m/s,

27.9 m/s2, (d) 12.6 m/s, 39.5 m/s; (f) -5.66 m]3. (Standard) [Comparing simple harmonic motion with uniform circular motion]A particle rotates counterclockwise in a circle of radius 3 m with a constant angular speed of 8 rad/s. At t=0, the particle has an x coordinate of 2 m and is moving to the right. Determine the x coordinate as a function of time.

[Ans.: ]If the particle were performing a simple harmonic motion along the x coordinate, what should be its angular frequency and initial phase to have the same x(t) dependence?

4. (Basic) If the amplitude of a simple harmonic oscillation is tripled, by what factor is energy changed ? [Ans: 9]

5. (Standard) Two mass-spring systems A and B oscillate so that their energies are equal. If kA=2kB, find the relation between the amplitudes of oscillation. [Ans: ]

6. (Standard) A block of mass 0.5 kg connected to a light spring for which the force constant is 100 N/m is free to oscillate on a horizontal, frictionless surface. It is given an initial velocity of 5 m/s at x=0 (equilibrium point). Find:

(a) the total energy of the body;(b) the amplitude of oscillations;(c) the velocity when the displacement is half the amplitude;(d) the displacement when the velocity is half the initial velocity;(e) the displacement when the kinetic and potential energy are the same;(f) the frequency and period of the motion;

30

Plot the displacement, velocity, and acceleration against time for one period.[Ans: (a) 6.25 J; (b) 0.35 m; (c) ±4.34 m/s; (d) ±0.31 m; (e) ±0.25 m; (f) 2.25 Hz, 0.44s]

7. (Standard) The shock absorbers in an old car with mass 1000 kg are completely worn out. When a 980-N person climbs slowly into the car to its centre of gravity, the car sinks 2.8 cm. When the car, with the person aboard, hits a bump, the car starts oscillating up and down in SHM. Model the car and person as a single body on a single spring, and find the period and frequency of oscillation. [Ans: 1.11 s; 0.9 Hz]

8. (Standard) An apple weighs 1 N. When you hang it from the end of a long spring of force constant 1.5 N/m and negligible mass, it bounces up and down in SHM. If you stop the bouncing and let the apple swing from side to side through a small angle, the frequency of this simple pendulum is half the bounce frequency. (Because the angle is small, the back and forth swingings do not cause any appreciable change in the length of the spring.) What is the unstretched length of the spring? [Ans: 2m]

9. (Standard) A simple pendulum consists of a 5.0000 kg mass suspended by 2.0000 m wire. It is pulled 10.0000 cm from the vertical and released. Find the maximum velocity of the pendulum to 5 significant figures, using g=9.8000 m/s2:(a) exactly (from the energy conservation law), and(b) making the small oscillation approximation (i.e. using ).

[Ans: (a) 0.22142 m/s; (b) 0.22136 m/s]

10. (Challenging) If we attach two blocks of masses m1 and m2 to either end of a spring of spring constant k and set them into oscillation, show that the oscillation frequency , where =m1m2/(m1+m2) is the reduced mass of the system.

11. (Standard) One of the vibrational modes of the HCl molecule has a frequency =8.9691013 rad/s. Using the relation stated in the previous problem, find the “spring constant” for HCl molecule. [Ans: 13.1 N/m]

31