15.0 Preview

As the name suggests, this chapter revolves around the study of waves and their propagation. We describe wave as a form of disturbance propagated from one region of space to another. We shall classify waves based on several criteria as described:

Waves which require a medium for propagation are called mechanical and those waves which do not require a medium is called non-mechanical or electro-magnetic waves; Waves that travel in a straight line are said to be one-dimensional, those which travel on a surface are two-dimensional and those that travel in space are three-dimensional; Waves in which the oscillations of particles of a medium are perpendicular to direction of wave propagation are called transverse waves. Waves in which the oscillations of particles of a medium are along the direction of wave propagation are called longitudinal waves.

We shall then state the expression to describe the displacement for an element of a wave as a sine function. We will define some of the important terminologies associated with this waveform: A point of maximum positive displacement (or the peak) in a wave is called crest, and a point of maximum negative displacement (or fall) is called trough; Amplitude is defined as the maximum displacement of a particle of the medium from its equilibrium position, as the wave propagates through the medium; wavelength of a wave is the least distance between any two consecutive troughs or crests; or between two consecutive points in the same phase of wave motion; The period of oscillation, T of a wave is defined as the time any string element takes to move through one complete oscillation; Frequency is the number of oscillations per unit time made by a string element as the wave passes through it.

We shall discuss about the principle of superposition as the phenomenon in which, two or more waves of the same nature travel past a point simultaneously, resulting in net disturbance, at the point, as the vector sum of disturbances due to the individual waves.

We will then study about the reflection of waves and discuss a special type of reflection, Echo. When a pulse or a travelling wave encounters a rigid body, the wave gets reflected. The sound heard after the incident sound wave gets reflected by a distant rigid boundary such as, a wall or a building is called, echo.

If two waves of equal wavelength are made to travel in opposite directions along a line, the waves superpose themselves under suitable conditions and produce standing waves. You will be interested to learn about them since such a type of waveform can be thought of as a snapshot of a wave in progress! Thus, it can be used to find quantities like amplitude and wavelength.

We shall study about the modes of vibration in some detail. The mode of vibration with lowest frequency is called the fundamental mode or the first harmonic and for consecutively higher frequencies, we use the term second harmonic, third harmonic and so on.

Next, we will learn about the phenomenon of beats. You would be familiar with this term. For instance, the beats produced by the table or that produced when two keys of a piano are struck simultaneously. There is a wavering of sound intensity when two such waves of sound of nearly same frequencies and amplitudes travelling in the same direction superimpose on each other and this is called beats.

We shall end with one significant phenomenon called, Doppler Effect as the apparent change in pitch or frequency of the sound heard by the observer due to the relative motion between the source and the observer. We will study some specific cases of the Doppler Effect; for instance, cases in which the source or the observer is moving or is stationary.

15.1 Introduction

Energy is transferred in two ways: 1. By bulk transfer of matter; 2. By wave propagation.

In wave transfer, there is no actual movement of matter. Here, a form of disturbance is generated in some part of the medium and the same is propagated to other parts. The disturbance is caused due to repeated periodic vibrations of particles about their mean positions and is passed on from one to another, without the flow of matter.

Thus, a wave is a form of disturbance, propagated from one region of space to another and this movement is called wave motion.

Mechanical and non-mechanical waves: Waves which require a medium for propagation are called mechanical and those waves which do not require a medium is called non-mechanical or electro-magnetic waves. Examples for electro-magnetic waves: Ultra-violet rays, X- rays, microwaves and radio waves; that generally travel with a speed of 3 x $10^8$ $ms^{-1}$ in vacuum.

Based on the way the waves spread, they are classified into three types:

1. One dimensional waves: Waves which travel along a straight line. For example: waves in a stretched string.
2. Two dimensional waves: Waves which travel on a surface. For example: waves on surface of water.
3. Three dimensional waves: Waves which travel in space. For example: Sound waves in air.

Questions for section 15.1:

1. What are waves? What are the two types of waves classified based on medium of propagation.
2. What are the three types of waves, classified based on how they spread? Give one example for each.

15.2 Transverse and Longitudinal Waves

Transverse Waves

• Waves in which the oscillations of particles of a medium are perpendicular to direction of wave propagation are called transverse waves.
• Transverse waves propagate in the form of crests and troughs.

Examples:

• Waves created on the surface of water.
• Waves generated in a string.

Longitudinal Waves

• Waves in which the oscillations of particles of a medium are along the direction of wave propagation are called longitudinal waves.
• Longitudinal waves are formed by a sequence of alternate compressions and rarefactions.

Examples:

• Sound waves in air are longitudinal.
• Waves set up in air columns (when the piston in a cylinder is pushed and pulled in a simple harmonic manner).

Note box: It is found that generally transverse and longitudinal waves travel with different speeds in the same medium.

The waves on the surface of water are of two kinds: capillary waves and gravity waves;

• Capillary waves are ripples of fairly short wavelength—no more than a few centimetres.
  The restoring force that produces them is the surface tension of water.
• Gravity waves have wavelengths typically ranging from several metres to several hundred metres.
  The restoring force that produces these waves is the pull of gravity, which tends to keep the water surface at its lowest level.
  The oscillations of the particles in these waves are not just propagated on the surface, but extend with diminishing amplitude to the very bottom. Thus, the particle motion in the water waves involves a complicated motion; they not only move up and down but also back and forth. The waves in an ocean are one such example of combination of both longitudinal and transverse waves.

Example 15.1: Given below are some examples of wave motion. State in each case if the wave motion is transverse, longitudinal or a combination of both:
(a) Motion of a kink in a longitudinal spring produced by displacing one end of the spring sideways.
(b) Waves produced in a cylinder containing a liquid by moving its piston back and forth.
(c) Waves produced by a motorboat sailing in water.
(d) Ultrasonic waves in air produced by a vibrating quartz crystal.

Solution: (a) Transverse and longitudinal
(b) Longitudinal
(c) Transverse and longitudinal
(d) Longitudinal

15.3 Displacement Relation in a Progressive Wave

To describe the propagation of a wave in a medium, we need a function that completely gives the shape of the wave at every instant of time. For example, to completely describe the wave on a string, we need a relation which describes the displacement of an element at a particular position as a function of time and also describes the state of vibration of various elements of the string along its length at a given instant of time.

Concept box: A sinusoidal wave is a mathematical curve that describes smooth repetitive oscillation. It is named after the sine-function.

• For a sinusoidal wave, as shown in Figure 15.1, particles of the wave perform simple harmonic vibrations.

Figure 15.1 A sinusoidal wave is sent along the string. A typical element of the string moves up and down continuously as the wave passes. It is a transverse wave.

• Let y (x, t) denote the transverse displacement of the element at position x, at time t.
• As the wave sweeps through succeeding elements of the string, the elements oscillate parallel to the y-axis. At any time t, the displacement y of the element located at position x is given by,
  y(x, t) = a sin (ωt – kx + φ) ………………. (15.2)
• We can also write it as a cosine function or a linear combination of sine and cosine functions such as,
  y(x, t) =A sin (ωt – kx) + B cos (ωt – kx)…………………… (15.3)
• Then, in Equation (15.2),
  $$a=\sqrt {A^2+B^2} \space and \space ∅= tan^{-1}⁡ \frac {B}{A} $$

Note box: Equation (15.2) represents a periodic function which, at any time t, gives the displacement of the elements of the string by virtue of their position. Equation (15.2) represents a progressive transverse wave travelling along the positive direction of the x-axis.

• On the other hand a function,
  y(x, t) = a sin (ωt + kx + φ )………………………. (15.4)
  like equation 15.4 represents a wave travelling in the negative direction of x-axis.

Introduction to some wave terminologies:
The following graphs represent plots of equation 15.2: y(x, t) = a sin (ωt – kx + φ) for five different values of “t”.

• A point of maximum positive displacement in a wave, shown by the arrow, is called crest, and a point of maximum negative displacement is called trough.
• The progress of the wave is indicated by the direction of the short arrow pointing to a crest of the wave (here, it is towards the right).

15.3.1 Amplitude and Phase

• Amplitude (a) of a wave is defined as the maximum displacement of a particle of the medium from its equilibrium position, as the wave propagates through the medium.
• The phase of the wave is the argument, “(ωt – kx + φ)” of the oscillatory term sin (ωt – kx + φ) in Equation (15.2).
  The phase of a wave describes the state of motion as the wave passes through a string element at a particular position, x. It changes linearly with time t.
• The sine function also changes with time, oscillating between +1 and –1.
  Its extreme positive value +1 corresponds to a peak of the wave moving through the element; at which the value of y at position x will be “a” (the amplitude).
  Its extreme negative value –1 corresponds to a valley (or fall) of the wave moving through the element, in which case, the value of y at position x will become “–a”.
• The constant, φ is called the initial phase angle. The value of φ is determined by substituting: t = 0 and by finding displacement and velocity of the element at x = 0.

15.3.2 Wavelength and Angular Wave Number

• The wavelength λ of a wave is the least distance between any two consecutive troughs or crests; or between two consecutive points in the same phase of wave motion.
• A typical wavelength is marked in Figure 15.3(a) above, which is a plot of Equation (15.2): y(x, t) = a sin (ωt – kx + φ); for t = 0 and φ = 0. At this time the equation reduces to,
  y(x, 0) = a sin kx ……………………………. (15.5)
• By definition, the displacement, y is same at both ends of this wavelength, that is, at x = $x_1$ and at x = $x_1$ + λ. Thus, by Equation (15.2),
  a sin k$x_1$ = a sin k($x_1$ + λ)
  = a sin (k $x_1$ + k λ)
• This condition can only be satisfied when,
  k λ = 2πn
  Where, n = 1, 2, 3… Since λ is defined as the least distance between points with the same phase n =1. Thus,
  $$ k= \frac {2π}{λ}………… (15.6)$$
  Where, k is called the propagation constant or the angular wave number.

Note box: SI unit of k is radian per metre or rad $m^{–1}$. Clearly, it can also be written as $m^{-1}$.

15.3.3 Period, Angular Frequency and Frequency

Figure 15.4 shows a graph of the displacement, y, of Equation (15.2), y(x, t) = a sin (ωt – kx + φ) versus time, t at a certain position along the string, where, x = 0.

Definition box: The period of oscillation, T of a wave is defined as the time any string element takes to move through one complete oscillation.

A typical period is marked on the Figure 15.7 as the distance, T.

Expression for angular frequency:

• For a sinusoidal wave propagating along positive direction of X- axis, we have:
  y = A sin (ωt – kx)
  For the vibration of particle at x=0, we get, y = A sin ωt
• Since sine function repeats after every increase in angle, 2π, we get, ωT = 2π
  ω = 2π/T ………………… (15.7)
  “ω” is called the angular frequency of the wave, its SI unit is rad s–1.
• The frequency, v of a wave is defined mathematically as 1/T and is related to the angular frequency ω by,
  $$ v= \frac {1}{T}= \frac {ω}{2π} …………(15.8)$$

Definition box: Frequency is the number of oscillations per unit time made by a string element as the wave passes through it.

Frequency is usually measured in hertz; which is itself the SI unit.

• In Equation (15.2), the displacement function for a longitudinal wave is written as,
  s(x, t) = a sin (ωt – kx + φ) ………………… (15.9)
  Where, s(x, t) is the displacement of an element of the medium in the direction of propagation of the wave at position x and time t.

Note box: All quantities in equation (15.9) have the same meaning as in case of a transverse wave except that the displacement function y (x, t) is to be replaced by the function s(x, t).

Q and A:

Why do we not get any information about the shape of a wave from the equation: y (0, t) = a sin (–ω t)?

Since x = 0 here, the wave does not show any progress at this point and the equation simply represents the point of origin itself. Thus, we cannot determine its shape with this equation.

Example 15.2: A wave travelling along a string is described by, y(x, t) = 0.005 sin (80.0 x – 3.0 t), in which the numerical constants are in SI units (0.005 m, 80.0 rad m$^{–1}$, and 3.0 rad $s^{–1}$). Calculate (a) the amplitude, (b) the wavelength, and (c) the period and frequency of the wave. Also, calculate the displacement y of the wave at a distance x = 30.0 cm and time t = 20 s?

Solution: On comparing this displacement equation with Eq. (15.2),
y (x, t ) = a sin (kx – ωt ),
we find

(a) the amplitude of the wave is 0.005 m = 5 mm.

(b) the angular wave number k and angular frequency ω are
$k = 80.0 \space m^{–1}$ and ω = 3.0 s$^{–1}$
We then relate the wavelength λ to k through Eq. (15.6),
λ = 2π/k
$$ = \frac {2π}{80.0 \space m^{- 1}} $$
= 7.85 cm

(c) Now we relate T to ω by the relation
T = 2π/ω
$= \frac{2π}{3.0 \space s^{-1}} $
$= 2.09 \space s $ and frequency, v = 1/T = 0.48 Hz

The displacement y at x = 30.0 cm and time t = 20 s is given by
$$ y = (0.005 \space m) sin \space (80.0 × 0.3 – 3.0 × 20) $$
$$= (0.005 \space m) sin \space (–36 + 12π)$$
$$= (0.005 \space m) sin \space (1.699)$$
$$ = (0.005 \space m) sin \space (97^0) \simeq 5 \space mm $$

Questions for section 15.3:

1. What are transverse waves? How do these waves travel or propagate? Give two examples.

2. What are longitudinal waves? How do these waves travel or propagate? Give two examples.

3. Write a note on the two types of waves on surface of water.

4. Name the two quantities required to describe the wave on a string.

5. Give the equation for displacement of elements of a string (or) the equation that represents a progressive wave travelling along positive x-direction.

6. What is crest and trough of a wave? Explain with figure.

7. Define:

a) Amplitude of a wave.

b) Wavelength of a wave

8. What is the phase of a wave? Illustrate with necessary equation. What is its significance?

9. What does a value of +1 and -1 denote for the sine function of a progressive wave equation denote?

10. What does the symbol, “φ” in the sine function denote?

11. Arrive at the expression for propagation constant for a progressive wave.

12. Define period (T) of oscillation. Represent it on a waveform.

13. Arrive at the expression for angular frequency and frequency. Specify the SI unit for both.

15.4 The Speed of a Travelling Wave

• Let us monitor the propagation of a travelling wave in the positive direction of x. We find that an element of string at a particular position x moves up and down as a function of time but the waveform advances to the right.
• The displacement of elements of the string at two different instants of time “t” differing by a small time interval “Δt” is depicted in Figure 15.8 (the phase angle φ has been taken to be zero).

Figure 15.5 Progression of a harmonic wave from time t to t + Δt. Where Δt is a small interval. The wave pattern as a whole shift to the right. The crest of the wave (or a point with any fixed phase) moves right by the distance Δx in time Δt

• It is observed that, during this interval of time, the entire wave pattern moves by a distance Δx in the positive direction of x. Thus, the wave is travelling to the right, in the positive direction of x.
• The ratio Δx/Δt is the wave speed v.
• As the wave moves (see Figure 15.8), each point of the moving waveform will have a different phase of the wave and displacement, y remains the same.
• Let us consider a point like, “A” marked on the peak of the waveform. If a point like, A on the waveform retains its displacement as it moves, it follows from Equation (15.2): y(x, t) = a sin (ωt – kx + φ) that, this is possible only when the argument is constant. Hence,
  ωt – kx = constant ……………………. (15.10)

Note box: Both x and t are changing; therefore, to keep the argument constant, if t increases, x must also increase. This is possible only when the wave is moving in the positive direction of x.

• To find the wave speed v, let us differentiate Equation (15.10) with respect to time;
  $$ \frac {d}{dt}(kx)- \frac {d}{dt}(ωt)=0 $$
  Making use of equations, 15.6 to 15.8; that is, $ k= \frac {2π}{λ} $ and $ ϑ= \frac {1}{T}= \frac {ω}{2π}$ respectively; we get,
  $$ k. \frac {dx}{dt}= ω=kv $$ $$ \frac {dx}{dt}= ν ………(15.11)$$ $$ ϑ= \frac {ω}{k}= \frac {λ}{T}=λν……(15.12)$$
• Equation (15.11) is a general relation valid for all progressive waves.
• The speed of a wave is related to its wavelength and frequency by the Equation (15.12).

Note box: The speed of the wave also depends on properties of the medium—If a wave is to travel in a medium like air, water, steel, or a stretched string, it must cause the particles of that medium to oscillate as it passes through it. For this to happen, the medium must possess mass and elasticity. Therefore, the linear mass density (or mass per unit length, in case of linear systems like a stretched string) and the elastic properties determine how fast the wave can travel in the medium.

15.4.1 Speed of a Transverse Wave on Stretched String

The speed of transverse waves on a string depends on two factors:

(i) The linear mass density or mass per unit length, μ (μ = mass m of the string divided by its length l).

(ii) The tension T.

Note box: 1. Linear mass density, μ has a dimension of [$ML^{–1}$]. 2. The tension, T has the dimension of force — namely, [M L $T^{–2}$].

• Our goal is to combine μ and T in such a way as to generate “v” [dimension (L $T^{–1}$)].
• On solving as shown below, we find that, the ratio T/μ has the following dimension:
  $$ \frac {[MLT^{-2}]}{[ML^{-1}]}=[L^2T^{-2}] $$
• Therefore, if v depends only on T and μ , the relation between them must be;
  $$ v=C \sqrt { \frac {T}{μ}}……………(15.13)$$
• C is found to be unity. Thus,
  $$ v=C\sqrt { \frac {T}{μ}}……………(15.14)$$
  Equation (15.14) tells us that, the speed of a wave along a stretched ideal string depends only on the tension and the linear mass density of the string and does not depend on the frequency of the wave.
• The frequency of the wave depends on the source that generates the wave and from Equation (15.12$ ϑ= \frac {ω}{k}= \frac {λ}{T}=λν$, the wavelength is given by,
  $$ λ= \frac {ϑ}{v}……………(15.15)$$

Example 15.3: A steel wire 0.72 m long has a mass of 5.0 ×10$^{–3}$ kg. If the wire is under a tension of 60 N, what is the speed of transverse waves on the wire?
Solution: Mass per unit length of the wire,
$$ μ = \frac { 5.0×10^{−3} \space kg}{ 0.72 \space m} $$
$$ = 6.9 ×10^{–3} \space kg \space m^{–1} $$
Tension, T = 60 N
The speed of wave on the wire is given by

$$ v = \sqrt{\frac{T}{ μ}} = \sqrt{\frac{60 \space N}{6.9 \times 10^{-3}\space kg m^{-1}}} = 93 \space ms^{-1}$$

15.4.2 Speed of a Longitudinal Wave (Speed of Sound)

The property that determines the extent to which the volume of an element of a medium changes when the pressure on it changes, is the bulk modulus B, as we defined earlier,
$$ B=- \frac {\Delta P}{\Delta V/V} …..(15.16)$$

Here ΔV/V is the fractional change in volume produced by a change in pressure ΔP. The SI unit for pressure is N $m^{–2}$ or pascal (Pa). Now since the longitudinal waves in a medium travel in the form of compressions and rarefactions or changes in density, the inertial property of the medium, which could be involved in the process, is the density ρ. The dimension of density is [M $L^{–3}$].Thus, the dimension of the ratio B/ρ is,

$$ \frac {[ML^{-1}T^{-2}]}{[ML^{-3}]}=[L^2T^{-2}]$$

Therefore,on the basis of dimensional analysis the most appropriate expression for the speed of longitudinal waves in a medium is $ v= C \sqrt{\frac {B}{ρ}} ….(15.18)$

where C is a dimensionless constant and can be shown to be unity. Thus the speed of longitudinal waves in a medium is given by,
$ v= \sqrt{\frac{B}{ρ}} ….(15.19)$

The speed of propagation of a longitudinal wave in a fluid therefore depends only on the bulk modulus and the density of the medium.When a solid bar is struck a blow at one end, the situation is somewhat different from that of a fluid confined in a tube or cylinder of constantcross section. For this case, the relevant modulus of elasticity is the Young’s modulus, since the sideway expansion of the bar is negligible and only longitudinal strain needs to be considered. It can be shown that the speed of a longitudinal wave in the bar is given by,
$v= \sqrt { \frac {Y}{ ρ}} ……(15.20)$

where Y is the Young’s modulus of the material of the bar.

Note box: It may be noted that, although the densities of liquids and solids are much higher than those of the gases, the speed of sound in them is higher. It is because, liquids and solids are less compressible than gases, and hence, they have much greater bulk modulus.

• As we know, for an ideal gas, the relation between pressure, P and volume, V is given by,
  PV = N$k_B$T …….. (15.21)
  where N is the number of molecules in volume V, $k_B$ is the Boltzmann constant and T the
  temperature of the gas (in Kelvin). Therefore, for an isothermal change it follows from Eq.(15.21)
  that VΔP + PΔV = 0
  $$ or \space – \frac {\Delta P}{\Delta V/V} =P $$
  Hence, substituting in Eq. (15.16), we have
  B = P
  Therefore, from Eq. (15.19) the speed of a longitudinal wave in an ideal gas is given by,
  $$ v= \sqrt{\frac {P}{ρ}} ….(15.22)$$
• This relation was first given by Newton and came to be known as Newton’s formula.

Example 15.4: Estimate the speed of sound in air at standard temperature and pressure. The mass of 1 mole of air is 29.0 ×10$^{–3}$ kg.

Solution: We know that 1 mole of any gas occupies 22.4 litres at STP. Therefore, density of air at STP is:
$ρ_o$ = (mass of one mole of air)/ (volume of one mole of air at STP)

$$= 1.29 \space kg \space m^{–3} $$

According to Newton’s formula for the speed of sound in a medium, we get for the speed of sound in air at STP,

$$ v = \bigg[ \frac{ 1.01×10^5 N m^{-2}}{1.29 \space kg \space m^{-3}} \bigg]^{1/2} = 280 \space ms^{-1} $$

Note box: The result shown in Equation (15.23) is about 15% smaller as compared to the experimental value of 331 m $s^{–1}$ as given in Table 15.1. If we examine the basic assumption made by Newton that, the pressure variations in a medium during propagation of sound are isothermal, we find that, this is not correct.

Laplace’s Correction:

• It was pointed out by Laplace that the pressure variations in the propagation of sound waves are so fast that there is little time for the heat flow to maintain constant temperature. These variations, therefore, are adiabatic and not isothermal.
• For adiabatic processes, the ideal gas satisfies the relation,

$$ PV^γ = constant$$
$$ i.e. Δ(PV^γ ) = 0$$
$$ or \space Pγ V ^{γ –1} ΔV + V^γ ΔP = 0$$

Thus for an ideal gas the adiabatic bulk modulus is given by,
$$ B_{ad}=- \frac {\Delta P}{\Delta V/V} $$
= γ P
where γ is the ratio of two specific heats,$C_p/C_v$. The speed of sound is, therefore, given by,
$$ v = \sqrt { \frac {γP}{ρ} } …..(15.24)$$

• This modification of Newton’s formula is referred to as the Laplace correction.

Note box: For air, γ = 7/5. Now, using Equation (15.24) , $ =\sqrt { \frac {γP}{ρ} }$ to estimate the speed of sound in air at STP, we get a value 331.3 m s–1, which agrees with the measured speed.

Questions for section 15.4:

1. Establish the relationship between speed of the wave, wavelength and its frequency.
2. Arrive at the expression that is used to find the speed of a transverse wave in a stretched string, in the form: $ v=\sqrt { \frac {T}{μ}}$
3. Arrive at the expression for speed of a longitudinal wave, in the form: $ v=\sqrt { \frac {P}{ρ} }$
4. Explain Laplace’s correction with relevant equations.

15.5 The Principle of Superposition of Waves

Definition box: Principle of superposition: When two or more waves of the same nature travel past a point simultaneously, the net disturbance, at the point is the vector sum of disturbances due to the individual waves.

Explanation:

• Let us consider that, two waves are travelling simultaneously along the same stretched string in opposite directions.
• The sequence of pictures shown in Figure 15.9 depicts the state of displacement of various elements of the string at different time instant.

Figure 15.6 Two pulses having equal and opposite displacements moving in opposite directions. The overlapping pulses add up to zero displacement in curve (c).

• Each picture depicts the resultant waveform in the string at a given instant of time. It is observed that, the net displacement of any element of the string at a given time is the algebraic sum of the displacements due to each wave. This way of addition of individual waveforms to determine the net waveform is called the principle of superposition.
• To put this rule in a mathematical form, let y1(x, t) and y2(x, t) be the displacements that any element of the string would experience if each wave travelled alone. The displacement y (x,t) of an element of the string when the waves overlap is then given by,
  $$ y (x, t) = y_1(x, t) + y_2(x, t) ……………(15.25) $$

The principle of superposition can also be expressed by stating that overlapping waves algebraically add to produce a resultant wave (or a net wave). The principle implies that the overlapping waves do not, in any way, alter the travel of each other.If we have two or more waves moving in the medium the resultant waveform is the sum of wave functions of individual waves. That is, if the wave functions of the moving waves are

$$ y_1 = f_1(x–vt),$$

$$y_2 = f_2(x–vt),$$
……….
……….

$y_n = f_n (x–vt)$ then the wave function describing the disturbance in the medium is
$ y = f_1(x – vt)+ f_2(x – vt)+ …+ f_n(x – vt) $

$$ = \Sigma_{i=1}^{n} f_i (x-vt) ….(15.26)$$

As illustrative examples of this principle we shall study the phenomena of interference and reflection of waves.
Let a wave travelling along a stretched string be given by,
$ y_1(x, t) = a sin (kx – ωt) ……. (15.27)$ and another wave, shifted from the first by a phase φ,
$ y_2(x, t) = a sin (kx – ωt + φ ) …….(15.28)$

Both the waves have the same angular frequency, same angular wave number k (same wavelength) and the same amplitude a.They travel in the positive direction of x-axis, with the same speed. Their phases at a given distance and time differ by a constant angle φ. These waves are said to be out of phase by φ or have a phase difference φ.
Now, applying the superposition principle, the resultant wave is the algebraic sum of the two constituent waves and has displacement
y (x, t ) = a sin (kx – ωt) + a sin (kx – ωt + φ) (15.29)

We now use the trigonometric relation

$$ sin \space α + sin \space β = 2 \space sin \frac {1}{2} (α + β)cos \frac {1}{2}(α – β) …..(15.30)$$

Applying this relation to Eq. (15.29) we have $ y (x, t) = [2a \space cos \frac {1}{2}φ]sin(kx – ωt + \frac {1}{2} φ)……..(15.31)$

Equation (15.31) shows that the resultant wave is also a sinusoidal wave, travelling in thepositive direction of x-axis.

The resultant wave differs from the constituent waves in two respects: (1) its phase angle is ($\frac{1}{2}$)φ and (2) its amplitude is the quantity in brackets in Eq. (15.31) viz.,

A(φ) = 2a cos ($\frac{1}{2}$)φ (15.32)

If φ = 0, i.e. the two waves are in phase,

Eq. (15.31) reduces to

A(0) = 2a sin (kx – ωt) (15.33)

The amplitude of the resultant wave is 2a, which is the largest possible value of A(φ).
If φ = π, the two waves are completely out of phase, the amplitude of the resultant wave given by Eq. (15.32) reduces to zero. We then have for all x and t,
y (x, t ) = 0 (15.34)

These cases are shown in figure 15.7 below,

Figure 15.7 The resultant of two harmonic waves of equal amplitude and wavelength according to the principle of superposition. The amplitude of the resultant wave depends on the phase difference φ, which is zero for (a) and π for (b).

15.6 Reflection of Waves

Echo:

• When a pulse or a travelling wave encounters a rigid body, the wave gets reflected. Echo is one such example of the reflection of sound waves from a rigid boundary. The sound heard after the incident sound wave gets reflected by a distant rigid boundary such as, a wall or a building is termed echo.
• To be precise, any sound we hear persists for 0.1 second even after the source stops producing sound. Taking the speed of sound in air to be 340 ms-1 or that sound travels 34m in 0.1 second is air. That means echo is heard if the rigid boundary reflecting sound is at a distance of 17m away from us.
• If the boundary is not completely rigid or is an interface between two different elastic media, a part of the wave is reflected and a part is transmitted into the medium, after which refraction occurs.
• Refraction is the change in direction of waves that occurs when waves travel from one medium to another. The waves after undergoing refraction are called, refracted waves.

Note box: The incident and refracted waves obey Snell’s law of refraction and also obey the usual laws of reflection.

Illustrations for reflection of waves:

To illustrate the reflection of waves at a boundary, we will consider two situations:

In first case, a string is fixed to a rigid wall at its left end, as shown in Figure 15.11(a). In the second case, the left end of the string is tied to a ring, which slides up and down without any friction on the rod, as shown in Figure 15.7(b). A pulse is allowed to propagate in both these strings. The pulse, on reaching the left end, gets reflected. The state of disturbance in the string at various times is shown in Fig. 15.7

Case 1:

• In Figure 15.11(a), the string is fixed to the wall at its left end. When the pulse arrives at this end, it exerts an upward force on the wall. By Newton’s third law, the wall exerts an opposite force of equal magnitude on the string. This second force generates a pulse at the support (the wall), which travels back along the string in the direction opposite to that of the incident pulse.
• In a reflection of this kind, there must be no displacement at the point of support as the string is fixed there. We can say that, the reflected and incident pulses must have opposite signs, so as to cancel each other at that point.
• Thus, in case of a travelling wave, the reflection at a rigid boundary will take place with a phase reversal or with a phase difference of π or 1800.

Case 2:

• In Figure 15.7(b), the string is fastened to a ring, which slides without friction on the rod. In this case, when the pulse arrives at the left end, the ring moves up the rod.
• As the ring moves, it pulls the string, stretching the string and producing a reflected pulse with the same sign and amplitude as the incident pulse. Thus, in such a reflection, the incident and reflected pulses get summed up, creating the maximum displacement at the end of the string: the maximum displacement of the ring (on receiving the incident ray) is twice the amplitude of either of the pulses. Thus, the reflection is without any additional phase shift.
• This case can be extended to a travelling wave; the reflection at an open boundary, such as the open end of an organ pipe, the reflection takes place without any phase change.
  We can thus, summarise the reflection of waves at a boundary or interface between two media as follows:
• A travelling wave moving toward a rigid boundary or a closed end is reflected with a phase reversal but the reflection at an open boundary takes place without any phase change.
• To express the above statement mathematically, let the incident wave be represented by
  $$ y_i(x, t) = a \space sin ( ),$$
  then, for reflection at a rigid boundary the reflected wave is represented by,
  $$ y_r(x, t) = a \space sin (kx + ω + π).$$
  = – a sin (kx + ωt) (15.35)
  For reflection at an open boundary, the reflected wave is represented by
  $$ y_r (x, t) = a \space sin (kx + ωt)……….. (15.36)$$

15.6.1 Standing Waves and Normal Modes

Introduction:

If two waves of equal wavelength are made to travel in opposite directions along a line, the waves superpose themselves under suitable conditions and produce standing waves. Standing waves can be used to produce a “still” effect or a snapshot (still picture) of wave and hence the quantities like amplitude and wavelength can be measured.

Formation of Stationary Waves— Theory

• Let us consider two progressive waves of same amplitude and wavelength travelling at the same speed along a line in opposite directions.
• Let the waves be represented by,
  $y_1$(x, t) = a sin (ωt – kx) (wave travelling in the positive direction of x-axis)
  $y_2$(x, t) = a sin (ωt + kx) (wave travelling in the negative direction of x-axis)
• The waves interfere to form a stationary wave. The equation for the resultant stationary wave is obtained using the principle of superposition. If “y” is the resultant displacement then,
  y(x, t) = $y_1$(x, t) + $y_2$(x, t)
  = a sin (kx – ωt) + a sin (kx + ωt)
  = (2a sin kx) cos ωt ……………. (15.37)
• The wave represented by Equation (15.37) does not describe a travelling wave, as the waveform or the disturbance does not move to either side.
• Here, the quantity, 2a sin kx within the brackets is the amplitude of oscillation of the element of the string located at the position x. In a travelling wave, in contrast, the amplitude of the wave is the same for all elements located at any point along the wave.
• Equation (15.37), therefore, represents a standing wave, a wave in which the waveform does not move. The formation of such waves is illustrated in Figure 15.9.

Figure 15.9 Stationary waves arising from superposition of two harmonic waves travelling in opposite directions. Note that the positions of zero displacement (nodes) remain fixed at all times.

• As we see, the points of maximum or minimum amplitude stay at one position in all of these cases. The amplitude is zero for values of kx that give sin kx = 0. The values for which sin kx = 0 are:
  kx = nπ, for n = 0, 1, 2, 3, …
  Substituting k = 2π/λ in this equation, we get,
  $$ x = n \frac {λ}{2} , for \space n = 0, 1, 2, 3, … (15.38)$$
  The positions of zero amplitude are called nodes. Note that a distance of $ \frac {λ}{2}$ or half a wavelength separates two consecutive nodes.The amplitude has a maximum value of 2a,which occurs for the values of kx that give|sin k x|= 1. Those values are
  kx = (n + ½) π for n = 0, 1, 2, 3, …
  Substituting k = 2π/λ in this equation, we get
  $$ x = (n + ½) \frac {λ}{2} , for \space n = 0, 1, 2, 3, … (15.39)$$
  as the positions of maximum amplitude. These are called the antinodes. The antinodes are separated by λ/2 and are located half way between pairs of nodes.
• For a stretched string of length L, fixed at both ends, the two ends of the string have to be nodes. If one of the ends is chosen as position x = 0, then the other end is x = L. In order that this end is a node; the length L must satisfy the condition;
  $$ L = n \frac {λ}{2}, for \space n = 1, 2, 3, … (15.40)$$
  This condition shows that standing waves on a string of length L have restricted wavelength given by
  $$ λ = \frac {2L}{n}, for \space n = 1, 2, 3, … etc…. (15.41)$$
  The frequencies corresponding to these wavelengths follow from Eq. (15.12) as
  $$ v = n \frac {v}{2L}, for \space n = 1, 2, 3, … etc…. (15.42)$$
  Where, v is the speed of travelling waves on the string and the set of frequencies given by Equation (15.42) are called the natural frequencies or modes of oscillation of the system.
• This equation tells us that, the natural frequencies of a string are integral multiples of the lowest frequency, $ ϑ= \frac {v}{2L} $ which corresponds to n = 1.

Characteristics of a standing wave can be summarized as below:

1. The waveform does not advance through a medium.
2. At certain positions in the medium, the particles are permanently at rest and the resultant displacement is zero. Such points are called nodes. The particles midway between the two adjacent nodes have maximum displacement and such positions are called antinodes.
3. All particles lying between two adjacent nodes move in the same direction at the same time. In other words, they have the same phase.
4. There is no flow of energy between any two sections of the medium.
5. The amplitude of vibration varies from point to point.
6. The waveform between two consecutive nodes is called a loop or a segment. The particles of the medium in a segment vibrate in same phase. The particles of one segment have a phase difference of π with particles of neighbouring segments.
7. The distance between any two successive nodes is $ \frac {λ}{2}$ and also the distance between any two successive antinodes is $ \frac {λ}{2}$ , where is the wavelength of the component wave. Thus, the length of one segment is equal to $ \frac {λ}{2}$ .

Stationary waves in air columns:

A column of air enclosed in a pipe can be set into vibrations by blowing air across its mouth or by holding a vibrating tuning fork near its mouth. Then, the longitudinal waves travel through the pipe and get reflected at the other end. The reflected waves move backwards. Thus, the incident and the reflected waves superpose. Longitudinal stationary waves are set up in air column as a result. At the closed end of the pipe, the wave particles are not free to vibrate and therefore, a node is formed. At the open end, on the other hand, the particles have maximum freedom for movement and hence, an antinode is formed here.

• If a pipe is closed at one end and open at the other end, it is called as closed pipe. On the other hand, if it is open at both ends, it is called an open pipe.
• An air column in a closed pipe or an open pipe can vibrate in a number of ways, giving rise to different pitches. The mode of vibration with lowest frequency is called the fundamental mode or the first harmonic. The second harmonic is the oscillation mode with n = 2 (In equation: $=n \frac {v}{2L}$). The third harmonic corresponds to n = 3 and so on. The frequencies associated with these modes are often labelled as $ϑ_1, ϑ_2, ϑ_3$ and so on. The other mods of vibration that produce notes of higher frequency, called overtones.
• The collection of all possible modes is called the harmonic series and “n” (in, $=n \frac {v}{2L}$) is called the harmonic number.

Modes of vibration of air column in a closed pipe:

• Let L be the length of a closed pipe and v be the velocity of sound in air. The air column in a closed pipe vibrates in such a way that always, antinode is formed at the open end and node at the closed end (regardless of the mode of vibration).

• In the first or fundamental mode of vibration, a node (N) is formed at the closed end and an antinode (AN) is formed at the open end as shown in figure (15.10 A). Therefore, the entire length L of the pipe comprises of half a segment. Let $v_1$ be the frequency and $λ_1$ be the wavelength of the fundamental note. Then,

$$ L = \frac {1}{2} \space segement = \frac {1}{2} \frac { λ_1}{2} = \frac { λ_1}{4} \space or \space λ_1= 4L $$
$$ ∴ ν _1 = \frac {v}{ λ_1}=\frac {v}{4L} …..(1) $$

This frequency $ ν _1$ is the fundamental frequency.

In the second mode of vibration

As shown in fig 15.10 (b),the length L of the pipe comprises of 1.5 segments. Let $ν _2$ be the frequency and $ λ_2$ be the wavelength of the first overtone. Then L=1.5 segment = $ \frac {3}{2} = \frac { λ_2}{2}$ = \frac {3}{4} λ_2$ or $ λ_2=\frac {4}{3} L $

$$ ∴ ν _2 = \frac {v}{ λ_2}= \frac {v}{\frac {4L}{3}}=\frac {3v}{4L} …(2)$$

This frequency is the third harmonic.

Similarly in the third mode of vibration, the length L of the pipe comprises of 2.5 segments as shown in figure 15.5 (c ). Let $ ν _3$ be the frequency and $ λ_3 $ be the wavelength of the second overtone, Then

$$ L=2.5 \space segments= \frac {5}{2} \frac { λ_3}{2}= \frac {5}{4} λ_3 \space or \space λ_3 = \frac {4}{5} L $$

$$ ∴ ν _3 = \frac {v}{ λ_3}= \frac {v}{\frac {4L}{5}}=\frac {5v}{4L} …(3)$$

This frequency is the fifth harmonic

From (1),(2) and (3),
$$ ν _1: ν _2: ν _3=1:3:5$$

• Therefore, in case of a closed pipe, the ratios of the frequency of the overtones to that of fundamental are odd natural numbers. In other words, only odd harmonics are present in an air column in a closed pipe.

Note box: The antinode at the open end of the pipe is not formed exactly at the end of the pipe but slightly beyond the end. This is because air molecules have maximum freedom to vibrate a little outside the open end and not just at the open end. This extra stretch beyond the pipe is the end correction in magnitude.

Modes of vibration of air column in an open pipe:

• Let L be the length of an open pipe and v be the velocity of sound in air. In the first harmonic or fundamental mode of vibration, two antinodes are formed at the open ends with a node in between. Therefore, the entire length, L of the pipe comprises of one segment as shown in figure, 15.11a.

• Let $ϑ_1$be the frequency and $λ_1$ be the wavelength of the fundamental note. Then,
  $$ L=1 \space segment = \frac { λ_1}{2} \space or \space λ_1 =2L $$
  $$ ∴ ν _1 = \frac{v}{ λ_1}=\frac {v}{2L} …..(1) $$

This frequency is the fundamental frequency.
In the second mode of vibration as shown in figure 15.6(b) the entire length L of the pipe comprises of 2 segments. Let $ ν _2 $ be the frequency and $ λ_2$ be the wavelength of the first overtone. Then,
$$ L=2 \space segment = 2 \frac { λ_2}{2} = λ_2 $$
$$ ∴ ν _2 = \frac {v}{ λ_2}=\frac {v}{L}=\frac {2v}{2L} …..(2) $$
This frequency is the second harmonic.
Similarly in the third mode of vibration the entire length L of the pipe comprises of 3 segments as shown in figure 15.6 (c ).Let $ ν _3$ be the frequency and $ λ_3$ be the wavelength of the second overtone. Then,
$$ L=3 \space segment = 3 \frac { λ_3}{2} = \frac {3}{2} λ_3 \space or \space λ_3 = \frac {2}{3}L $$
$$ ∴ ν _3 = \frac {v}{ λ_3}=\frac {v}{\frac {2L}{3}}=\frac {3v}{2L} …..(3) $$
This frequency is the third harmonic.
From (1),(2) and (3)
$$ ν _1: ν _2: ν _3=1:2:3$$

• Therefore, in case of an open pipe, the ratios of the frequency of the overtones to that of the fundamental are natural numbers. Thus, all harmonics are present for vibrations of an air column in an open pipe.

Modes of vibration in a stretched string;

• When a stretched string (or wire) vibrates, stationary waves are set up because of reflections at the two fixed ends. The direct and reflected waves superpose over each other giving rise to traverse stationary waves.
• The two fixed ends of the string are always nodes. The string vibrates in such a way that, it is divided into an integral number of equal segments (loops).
• Consider a uniform string of length, l stretched between two fixed points, A and B. If M is the mass of the string, then the mass per unit length or the linear mass density is $m = \frac {M}{l}$

• The velocity of propagation of transverse progressive wave along a stretched string is given by $ , v= \sqrt{\frac{T}{m}}$ where T is tension in the string.
• In the fundamental mode, the entire length, l of the string vibrates in one segment as shown in figure 15.12(a). Let $ϑ_1$ be the frequency and $λ_1$ be the wavelength of the fundamental note. Then,$$ l= \frac {λ_1}{2} \space or \space λ_1 = 2l $$
  $$ ∴ ν _1 = \frac {v}{ λ_1} = \frac {1}{2l} \sqrt { \frac {T}{m}} …..(1)$$

This frequency is the fundamental frequency ….(1)
In the second mode of vibration, the entire length of the string vibrates in two segements.[Fig.15.12(b)].Let $ ν _2 $ be the frequency and $ λ_2 $ be the wavelength of the first overtone. Then,

$$ l = \frac { 2λ_2}{2} = λ_2 $$
$$ ∴ ν _2 = \frac {v}{ λ_2} = \frac {1}{l} \sqrt { \frac {T}{m}} = \frac {2}{2l}\sqrt { \frac {T}{m}} …..(2)$$

This frequency is the second harmonic.

In the third mode of vibration, the entire length of the string vibrates in three segments [Fig.15.12 ( c )].Let $ ν _3$ be the frequency and $ λ_3 $ be the wavelength of the second overtone. Then
$$ l = \frac {3 λ_3}{2} \space or \space λ_3 = \frac {2l}{3} $$
$$ ∴ ν _3 = \frac {v}{ λ_3} = \frac {3}{2l} \sqrt { \frac {T}{m}} …..(3)$$

This frequency is the third harmonic.

From eq.(1),(2) and (3) $ ν _1: ν _2: ν _3=1:2:3$ .Therefore, all harmonies are present in the vibrations of a stretched string.

In general , if p segments are formed in length l of the string , then $ l = p \frac { λ_2}{2} $ or $ λ = \frac {2l}{p}$.

If ν is the frequency of the string, $ ν = \frac {v}{ λ} = \frac {p}{2l} \sqrt { \frac {T}{m}} $

If r is the radius of the wire and $ \rho $ is the density of its material, we get
$$ m = \frac {M}{l} = \frac { \pi r^2 l \rho}{l}= \pi r^2 \rho $$

The fundamental frequency of the string is
$$ i.e, ν = \frac {1}{2l} \sqrt { \frac {T}{m}} = \frac {1}{2l} \sqrt { \frac {T}{\pi r^2 \rho }}$$

Example 15.5: A pipe, 30.0 cm long, is open at both ends. Which harmonic mode of the pipe rasonates a 1.1 kHz source? Will resonance with the same source be observed if one end of the pipe is closed ? Take the speed of sound in air as 330 m s$^{–1}$

Solution:

The first harmonic frequency is given by
$$ν_1 = \frac {v}{λ_1} = \frac { v}{2L} \hspace{10mm} (open \space pipe) $$

where L is the length of the pipe. The frequency of its nth harmonic is:

$$ ν_n= \frac { nv}{2L} , for n = 1, 2, 3, … \hspace{10mm} (open \space pipe) $$

First few modes of an open pipe are shown in Fig. 15.14.
For L = 30.0 cm, v = 330 m s$^{–1}$,

$$ ν_n = \frac { n×330(ms^{-1})}{0.6 \space (m) } = 550 \space ns^{-1} $$

Clearly, a source of frequency 1.1 kHz will resonate at $v_2$, i.e. the second harmonic.

Now if one end of the pipe is closed (Fig. 15.15), it follows from Eq. (14.50) that the fundamental frequency is $ ν_1 = \frac {v}{λ_1} = \frac { v}{4L} \space (pipe \space closed \space at \space one \space end) $ and only the odd numbered harmonics are present :
$ ν_3 =\frac {3 v}{4L} $,$ ν_5= \frac {5 v}{4L}$ , and so on
For L = 30 cm and v = 330 m s$^{–1}$, the fundamental frequency of the pipe closed at one end is 275 Hz and the source frequency corresponds to its fourth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed.

Example 15.6: Two sitar strings A and B playing the note ‘Dha’ are slightly out of tune and produce beats of frequency 5 Hz. The tension of the string B is slightly increased and the beat frequency is found to decrease to 3 Hz. What is the original frequency of B if the frequency of A is 427 Hz ?

Solution: Increase in the tension of a string increases its frequency. If the original frequency of B ($νB$) were greater than that of A ($ν_A$ ), further increase in $ν_B$ should have resulted in an increase in the beat frequency. But the beat frequency is found to decrease. This shows that $ ν_B < ν_A$. Since $ν_A – ν_B$ = 5 Hz, and $ν_A$ = 427 Hz, we get$ νB$ = 422 Hz.

Example 15.7: A rocket is moving at a speed of 200 m s$^{–1}$ towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket.

Solution: (1) The observer is at rest and the source is moving with a speed of 200 m s$^{–1}$. Since this is comparable with the velocity of sound, 330 m s$^{–1}$, we must use Eq. (15.50) and not the approximate Eq. (15.51). Since the source is approaching a stationary target, $v_o$ = 0, and $v_s$ must be replaced by –$v_s$. Thus, we have

$$ v = v_0 \bigg( 1 – \frac {v_s}{ v} \bigg)^{-1} $$

$$ v = 1000 \space Hz × [1 – 200 \space m s^{–1}/330 \space m s^{–1}]^{–1} $$

$$ \simeq 2540 Hz $$

(2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, $v_s$ = 0 and $v_o$ has a positive value.

The frequency of the sound emitted by the source (the target) is v, the frequency intercepted by the target and not $v_o$. Therefore, the frequency as registered by the rocket is

$$ v′ = v \bigg( \frac{v+v_o}{v} $$

$$ =2540 \space Hz × \frac{ 200ms^{-1} +330ms^{-1}}{300 \space ms^{-1}} $$

$$ \simeq 4080 Hz $$

Questions for section 15.5 and 15.6

1. State and explain the principle of superposition of two waves.

2. Write a note on how echo occurs.

3. What is refraction?

4. Explain how waves get reflected back when they hit a rigid wall and when they hit an open boundary; with illustrations for each of the two cases.

5. What is a standing wave? Arrive at the expression for a standing wave in the form: y (x,t) = (2a sin kx) cos ωt

6. What are nodes and antinodes? Explain with necessary equations.

7. What do we mean by fundamental mode of vibration?

8. What are overtones?

9. What are harmonic series and a harmonic number?

10. Show that in a closed pipe, the ratios of frequency of overtones to that of fundamental frequency are odd natural numbers. Explain with figures.

11. Show that in an open pipe, the ratios of frequencies of overtones to that of fundamental are natural numbers. Explain with figure.

12. Arrive at the expression for fundamental frequency of a stretched string. Explain with figure.

15.7 Beats

Example box: If we listen, a few minutes apart, two sounds of very close frequencies, say 256 Hz and 260 Hz, we will not be able to distinguish between them. However, if both these sounds reach our ears simultaneously, what we hear is a sound of frequency 258 Hz, the average of the two combining frequencies. This is due to superposition of waves produced by two tuning forks.

• If the two sources of sound of slightly different frequencies are sounded together, the loudness of sound varies between maximum, called waxing and minimum, called waning periodically with time.
• This phenomenon of wavering of sound intensity when two waves of nearly same frequencies and amplitudes travelling in the same direction superimpose on each other, is called beats.
• For example, beats can be heard when two adjacent keys of a piano are struck simultaneously.
• The time interval between two consecutive waxing or two consecutive waning is called beat period.
• The number of beats heard per second is called beat frequency. In other words, number of waxing or waning heard per second is beat frequency.
• It is found that beat frequency is equal to the difference in frequencies of the two sound waves. If $ν _1$ and $ ν _2$ are the frequencies of the two producing beats, beat frequency = $ ν _1- ν _2$

Formation of beats:

• The formation of beats is illustrated in figure 15.13.

• Two sound waves with two different frequencies travelling in the same direction undergoing superposition are represented by (a) and (b).
• At certain instant of time, $t_1$, the displacements of a particle due to the two waves are in same direction. Hence, the resultant displacement is maximum.
• Due to the difference in periods, the two waves gradually go out of phase. At time, $t_2$ the displacements are in opposite directions and the resultant displacement is minimum (waning).
• At time $t_3$, the displacements are in same direction once again and the resultant displacement is maximum (waxing) and so on.
• The variation of the resultant displacement with time is represented by figure: c.

Theory of beats:

• Consider two sound waves of frequencies $ ν _1$ and $ ν _2$ propagating simultaneously in the same direction. For simplicity, we assume that, the amplitudes of two waves are equal at a location, x = 0. Then, the two waves are represented as,

$$ y_1=A \space sin \space ω_1 t \space and \space y_2=A \space sin\space ω_2 t $$

Where, $ ω_1=2πv_1$ and $ω_2=2πv_2$ are the angular frequencies of the two waves.

• The resultant displacement at a point due to superposition of the two waves is:

$$ y=y_1+y_2=A \space sinω_1 t+ A \space sinω_2 t$$

• The time interval between two consecutive minima is equal to$ \frac {1}{ ν _1- ν _2 }$. The beat period is the time required for one beat to be formed and it is equal to the time interval between two successive minima, as defined earlier. Thus,
  Beat Period = $\frac{1}{v_1 – v_2}$
• The number of beats formed per second is the beat frequency and it is equal to the reciprocal of beat period. Thus, beat frequency = $ ν _1- ν _2$
• Hence, the number of beats formed per second is equal to the differences between the frequencies of the two sound waves.

Questions for section 15.7:

1. What is waxing and waning?
2. Define:
   a) Beat
   b) Beat period
   c) Beat frequency
3. Explain the formation of beats with figures.
4. Arrive at the expression for the resultant wave when to waves superpose to form beats, in the form. Give the expressions for maximum and minimum amplitudes. Consequently, arrive at the expression for beat period.

15.8 Doppler Effect

It is a common experience that the pitch or the shrillness of sound heard by an observer (listener) changes when the source of sound and the observer are in relative motion. When a source of sound is moving towards a stationary observer, the pitch of the sound heard is higher than that when the source is at rest.

When the source of sound is moving away from the stationary observer, the pitch of the sound heard is lower than that when the source of sound is at rest. The vice versa is also true. When the source is at rest and the observer is moving away, the pitch of the sound is again lower.

Definition box: The apparent change in pitch or frequency of the sound heard by the observer due to the relative motion between the source and the observer is called Doppler Effect.

Example box: 1. For a person on a railways platform, the pitch of the whistle of a stationary railway engine heard will be the same as the pitch of the whistle emitted by the engine. When the engine moves towards the person, the pitch heard by the person appears to be higher and when the engine moves away from the person, the pitch of the whistle appears to be lower. But for the driver of the engine, the pitch heard will be same in either of the cases.   2. Similarly, when a car moves past a person on the roadside at high speed, the pitch of the horn for the person appears to be higher when the car approaches the person and the pitch of the horn appears to be lower when the car moves away from the person. But for the driver, the pitch of the horn heard is always the same.

Note box: Doppler Effect is common to all types of motions, including light!

Case 1: Doppler Effect when source is moving and observer is stationary

• Let us consider a source of sound, S moving towards the observer O at rest. Let v be the velocity of sound waves emitted by the source S, $v_s$ be the velocity of source moving toward the observer O, be the frequency of sound waves emitted by the source, S and λ be the wavelength of sound waves emitted by S.
• When the source is stationary, waves emitted by it in one second can be represented by SP = ν as shown in figure 15.14 (a).

• When the source, S is moving towards the stationary observer, O, it moves from S to S’ in one second; such that, SS’ = $v_s$; the first wave is emitted at S (not shown in figure) and the next at S’. Therefore, the next set of waves emitted occupy S’P = v – $v_s$ as shown in figure 15.14 (b).

• From the two figures, it is clear that the wavelength of the waves have changed. If λ’ is the apparent wavelength of the waves received by the observer,

$$ λ’ = \frac { (v- v_s)}{ λ} $$

As the velocity of the sound with respect to the observer remains the same, the apparent frequency ν’ of the waves received by the observer is given by
$$ ν’ = \frac {v}{ λ’}= ν \times \frac {v}{v-v_s} \space or \space ν’= ν \bigg( \frac {v}{v-v_s} \bigg) $$

Since $ \big( \frac {v}{v-v_s} \big) > 1 $ ,we get ν’ > ν

Thus ,when the source approaches a stationary observer, the apparent frequency ν’ is more than the actual frequency ν of the source.

In case the source is moving away from the observer , $v_s$ is taken to be negative and the apparent frequency is given by

$$ ν’ = ν \bigg( \frac {v}{v+v_s} \bigg) $$

• In this case, the apparent frequency ν’ is less than the actual frequency, ν of the source.

Case 2: Doppler Effect when observer is moving and source is stationary:

• Let us consider the observer moving towards the stationary source, S. Let v be the velocity of the sound waves emitted by the source, $v_0$ be the velocity of the observer and ν be the frequency and λ be the wavelength of the waves emitted by the source, S.
• As the source is stationary, the wavelength of the waves emitted by it remains unchanged. The source emits waves in one second and they occupy the distance SP = v as shown in figure 15.15 below;

These waves travel towards the observer, O with a velocity, $v_0$. Therefore, the sound waves and the observer move towards each other. Relative velocity of sound waves with respect to the observer is : $$ v’ = v + v_0$$

• Therefore, the apparent frequency of sound waves received by the observer is,

$$ ν’ = \frac{v’}{ λ} = \frac{(v + v_s)}{λ} $$

Since ν =v λ or $ λ = \frac {v}{ ν }$,we have $ ν’ = \frac {(v+v_o)}{\big( \frac {v}{ ν} \big)} $

$$ ∴ ν’ = ν \bigg( \frac {v+v_o}{v} \bigg) \space Since \space \bigg( \frac { v+v_o}{v} \bigg) > 1,we \space get \space ν’ > ν $$

Thus, when an observer approaches a stationary source the apparent frequency ν’ is more than the actual frequency ν.

In case the source is moving away from the stationary source , $v_o$ is taken as negative , and the apparent frequency is given by
$$ ν’= ν \bigg( \frac {v-v_o}{v} \bigg) $$

In this case, the apparent frequency ν’ is less than the actual frequency, ν of the source.

Case 3: Doppler Effect when source and observer, both are in motion.

(i) When the source and observer are moving toward each other:

• Let us consider a source, S emitting sound waves of frequency ν, and wavelength, $\lamda$ moving toward an observer, O with the velocity, $v_s$ as shown in figure 15.16a below:

• Let the observer, O move towards the source, S with a velocity, $v_0$.
• The source S, emits waves in one second. These waves travel with a velocity, v toward the observer, O. Since the source travels a distance, $v_s$ in one second, the waves occupy the length, $v – v_s$. Therefore, the apparent wavelength of the waves is,
• $$ λ’ = \frac {(v-v_s)}{ ν} $$

The waves with this modified wavelength λ’ move towards the observer with velocity v. But the observer is moving towards the source with velocity $v_o$. Therefore the relative velocity with which sound waves approaches the observer is $ v’=(v+v_o)$
Thus, the apparent frequency of sound waves received by the observer is
$$ ν’ = \frac {v’}{ λ’} = \frac { (v+v_o)}{ λ’}=(v+v_o) \times \frac { ν}{(v-v_s)} \space or \space ν’ = ν \bigg( \frac {v+v_o}{v-v_s} $$

(ii) When the source and observer are moving away from each other (Fig 15.11(b)

$ λ’ = \frac {(v+v_s)}{ ν} $ and relative velocity $v’=v-v_o$
$$ ∴ ν’= \frac { (v-v_o)}{ λ’}=(v-v_o) \times \frac { ν}{(v+v_s)} \hspace{10mm} i.e., ν’= ν \bigg( \frac {v-v_o}{v+v_s} \bigg) $$

Applications of Doppler Effect:

• The change in frequency caused by a moving object due to Doppler Effect is used to is used by police to check over-speeding of vehicles.
• It is used at airports to guide aircraft, and in the military to detect enemy aircraft.
• Astrophysicists use it to measure the velocities of stars.
• Doctors use it to study heart beats and blood flow in different part of the body. They use ulltrasonic waves, and in common practice, it is called sonography. Ultrasonic waves enter the body of the person, some of them are reflected back, and give information about motion of blood and pulsation of heart valves, as well as pulsation of the heart of the foetus. In the case of heart, the picture generated is called echocardiogram.

Example 15.7 A rocket is moving at a speed of 200 m $s^{–1}$ towards a stationary target. While moving, it emits a wave of frequency 1000 Hz. Some of the sound reaching the target gets reflected back to the rocket as an echo. Calculate (1) the frequency of the sound as detected by the target and (2) the frequency of the echo as detected by the rocket.

Solution:

(1) The observer is at rest and the source is moving with a speed of 200 m $s^{–1}$. Since this is comparable with the velocity of sound, 330 m $s^{–1}$, we must use Eq. (15.50) and not the approximate Eq. (15.51). Since the source is approaching a stationary target, $v_o$= 0, and $v_s$ must be replaced by –$v_s$. Thus, we have

$$ v = v_o \bigg( 1 – \frac {v_s}{v} \bigg)^{-1}$$

$$ v = 1000 \space Hz [1 – 200 \space ms^{-1}/330 \space m \space s^{-1}$$

$$ 2540 \space Hz$$

(2) The target is now the source (because it is the source of echo) and the rocket’s detector is now the detector or observer (because it detects echo). Thus, vs = 0 and vo has a positive value.

The frequency of the sound emitted by the source (the target) is v, the frequency intercepted by the target and not vo. Therefore, the frequency as registered by the rocket is $ v’ = v \bigg( \frac {v + v_o}{v} \bigg) $

$$ = 2540 \times \frac {200 \space ms^{-1} + 330 \space ms^{-1}}{330 \space ms^{-1}}$$
$$ 4080 \space Hz$$

Questions for section: 15.8

1. Define Doppler Effect. Give one example.
2. Explain Doppler Effect when the source is moving and the observer is stationary; with relevant figures and equations.
3. Explain Doppler Effect when the observer is moving and the source is stationary; with a relevant figure and equations.
4. Explain Doppler Effect when the observer and the source are moving towards each other; with a relevant figure and equations.
5. Explain Doppler Effect when the observer and the source are moving away from each other; with a relevant figure and equations.
6. List the applications of Doppler Effect.