# Physics NEET: Waves – Sound and Light

## Course Content

Total learning: 2 lessons / 1 quiz

### Introduction

In the previous chapter ‘Oscillations’, we studied the motion of objects oscillating in isolation. What happens in a system, which is a collection of such objects? A material medium provides such an example. Here, elastic forces bind the constituents to each other and, therefore, the motion of one affects that of the other. If you drop a little pebble in a pond of still water, the water surface gets disturbed. The disturbance does not remain confined to one place, but propagates outward along a circle. If you continue dropping pebbles in the pond, you see circles rapidly moving outward from the point where the water surface is disturbed. It gives a feeling as if the water is moving outward from the point of disturbance. If you put some cork pieces on the disturbed surface, it is seen that some cork pieces move up and down but do not move away from the centre of disturbance. This shows that the water mass does not flow outward with the circles, but rather a moving disturbance is created. Similarly, when we speak, the sound moves outward from us, without any flow of air from one part of the medium to another. The disturbances produced in air are much less obvious and only our ears or a microphone can detect them. These patterns, which move without the actual physical transfer or flow of matter as a whole, are called waves. In this chapter, we will study such waves.

Waves transport energy and the pattern of disturbance has information that propagate from one point to another. All our communications essentially depend on transmission of signals through waves. Speech means production of sound waves in air and hearing amounts to their detection. Often, communication involves different kinds of waves. For example, sound waves may be first converted in to an electric current signal which in turn may generate an electromagnetic wave that may be transmitted by an optical cable or via a satellite. Detection of the original signal will usually involve these steps in reverse order.

Not all waves require a medium for their propagation. We know that light waves can travel through vacuum. The light emitted by stars, which are hundreds of light years away, reaches us through inter-stellar space, which is practically vacuum.

The most familiar type of waves such as waves on a string, water waves, sound waves, seismic waves, etc., is the so-called mechanical waves. These waves require a medium for propagation, they cannot propagate through vacuum. They involve oscillations of constituent particles and depend on the elastic properties of the medium. The electromagnetic waves are a different type of wave. Electromagnetic waves do not require a medium – they can travel through vacuum. Light, radio waves, X-rays, are all electromagnetic waves. In vacuum, all electromagnetic waves have the same speed c, whose value is:

c = 299,792,458 ms-1 ……………………………………………..(1)

A third kind of wave is the so-called Matter waves. They are associated with constituents of matter: electrons, protons, neutrons, atom and molecules. They arise in quantum mechanical description of nature that you will learn in your later studies. Though conceptually more abstract than mechanical or electromagnetic waves, they have already found applications in several devices basic modern technology; matter waves associated with electrons are employed in electron microscopes.

In this chapter we will study mechanical waves, which require a material medium for their propagation.

The aesthetic influence of waves on art and literature is seen from very early times; yet the first scientific analysis of wave motion dates back to seventeenth century. Some of the famous scientists associated with the physics of wave motion are Christiaan Huygens (1629-1695), Robert Hooke and Isaac Newton. The understanding of physics of waves followed the physics of oscillations of masses tied to springs and physics of oscillations of masses tied to springs and physics of the simple pendulum. Waves in elastic media are intimately connected with harmonic oscillations. (Stretched strings, coiled springs, air, etc., are examples of elastic media).

We shall illustrate this connection through simple examples.

Consider a collection of springs connected to one another as shown in Fig. 1. If the spring at one end is pulled suddenly and released, the disturbance travels to the other end. What has happened? Fig 1 – A collection of springs connected to each other. The end A is pulled suddenly generating a disturbance, which then propagates to the other end.

The first spring is disturbed from its equilibrium length. Since the second spring is connected to the first, it is also stretched or compressed, and so on. The disturbance moves from one end to the other; but each spring only executes small oscillations about its equilibrium position. As a practical example of this situation, consider a stationary train at a railway station. Different bogies of the train are coupled to each other through a spring coupling. When an engine is attached at one end, it gives a push to the bogie to another with another without the entire train being bodily displaced.

Now let us consider the propagation of sound waves in air. As the wave passes through air, it compresses or expands a small region of air. This causes a change in the density of that region, say δp, this change induces a change in pressure, δp, in that region. Pressure is force per unit area, so there is a restoring force proportional to the disturbance, just like in a spring. In this case, the quantity similar to extension or compression of the spring is the change in density. If a region is compressed, the molecules in that region are packed together, and they tend to move out to the adjoining region, thereby increasing the density or creating compression in the adjoining region. Consequently, the air in the first region undergoes rarefaction. If a region is comparatively rarefied the surrounding air will rush in making the rarefaction move to the adjoining region. Thus, the compression or rarefaction moves from one region to another, making the propagation of a disturbance possible in air.

In solids, similar arguments can be made. In a crystalline solid, atoms or group of atoms are arranged in a periodic lattice. In these, each atom or group of atoms is in equilibrium, due to forces from the surrounding atoms. Displacing forces from the surrounding atoms. Displacing one atom, keeping the others fixed, leads to restoring forces, exactly as in a spring. So we can think of atoms in a lattice as end points, with springs between pairs of them.

In the subsequent sections of this chapter we are going discuss various characteristic properties of waves.

### Transverse and Longitudinal Waves

We have seen that motion of mechanical waves involves oscillations of constituents of the medium. If the constituents of the medium oscillate perpendicular to the direction of wave propagation, we call the wave a transverse wave. If they oscillate along the direction of wave propagation, we call the wave a longitudinal wave.

Fig. 2 shows the propagation of a single pulse along a string, resulting from s single up and down jerk. If the string is very long compared to the size of the pulse, the pulse will damp out before it reaches the other end and reflection from that end may be ignored. Fig 2 – When a pulse travels along the length of a stretched string (x-direction), the elements of the string oscillate up and down (y-direction)

Fig 3 shows a similar situation, but this time the external agent gives a continuous periodic sinusoidal up and down jerk to one end of the string. The resulting disturbance on the string is then a sinusoidal wave. In either case the elements of the string oscillate about their equilibrium mean position as the pulse or wave passes through them. The oscillations are normal to the direction of wave motion along the string , so this is an example of transverse wave. Fig 3 –

A harmonic (sinusoidal) wave travelling along a stretched string is an example of a transverse wave. An element of the string in the region of the wave oscillates about its equilibrium position perpendicular to the direction of wave propagation.

We can look at a wave in two ways. We can fix an instant of time and picture the wave in space. This will give us the shape of the wave as a whole in space at a given instant. Another way is to fix a location i.e., fix our attention on a particular element of string and see its oscillatory motion in time.

Fig. 4 describes the situation for longitudinal waves in the most familiar example of the propagation of sound waves. A long pipe filled with air has a piston at one end. A single sudden push forward and pull back of the piston will generate a pulse of condensations (higher density) and rarefactions (lower density) in the medium (air). If the push-pull of the piston is continuous and periodic (sinusoidal), a sinusoidal wave will be generated propagating in air along the length of the pipe. This is clearly an example of longitudinal waves. Fig 4 – Longitudinal waves (sound) generated in a pipe filled with air by moving the piston up and down. A volume element of air oscillates in the direction parallel to the direction of wave propagation.

The waves considered above, transverse or longitudinal, are travelling or progressive waves since they travel from one part of the medium to another. The material medium as a whole does not move, as already noted. A stream, for example, constitutes motion of water as a whole. In a water wave, it is the disturbance that moves, not water as a whole. Likewise a wind (motion of air as a whole) should not be confused with a sound wave which is a propagation of disturbance (in pressure density) in air, without the motion of air medium as a whole.

In transverse wave, the particle motion is normal to the direction of propagation of the wave. Therefore, as the wave propagates, each element of the medium undergoes a shearing strain. Transverse waves can, therefore, be propagated only in those media, which can sustain shearing stress, such as solids and not in fluids. Fluids, as well as, solids can sustain compressive strain; therefore, longitudinal waves can be propagated in all elastic media. For example, in medium like steel, both transverse and longitudinal waves can propagate, while air can sustain only longitudinal waves. The waves on the surface of water are of two kinds: capillary waves and gravity waves. The former are ripples of fairly short wavelength-not more than a few centimetre- and the restoring force that produces them is the surface tension of water. Gravity waves have lengths 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 involves a complicated motion-they not only move up and down but also back and forth. The waves in an ocean are the combination of both longitudinal and transverse waves.

It is found that, generally, transverse and longitudinal waves travel with different speed in the same medium.

Example 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 en 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 motor boat sailing in water.

(d) Ultrasonic waves in air produced by a vibrating quartz crystal.

(a) Transverse and longitudinal

(b) Longitudinal

(c) Transverse and longitudinal

(d) Longitudinal

### Displacement Relation in a Progressive Wave

For mathematical description of a travelling wave, we need a function of both position x and time t. Such a function at every instant should give the shape of the wave at that instant. Also, at every given location, it should describe the motion of the constituent of the medium at that location. If we wish to describe a sinusoidal travelling wave (such as the one shown in Fig. 3) the corresponding function must also be sinusoidal. For convenience, we shall take the wave to be transverse so that if the position of the constituents of the medium denoted by x, the displacement from the equilibrium position may be denoted by y. A sinusoidal travelling wave is then described by:

y(x,t) = a sin(kx- ωt + Φ) ………………………………………………………(2)

The term Φ in the argument of sine function means equivalently that we are considering a linear combination of sine and cosine functions:

y(x,t) = A sin(kx-ωt)+Bcos(kx-ωt)………………………………………………….(3)

From Equations (2) and (3),

a = √(A² + B²) and Φ = tan-1(B/A)

To understand why Equation (2) represents a sinusoidal travelling wave, take fixed instant, say t = t0. Then, the argument of the sine function Equation (2) is simply kx+constant. Thus, the shape of the wave (at any fixed instant) as a function of x is a sine wave. Similarly, take a fixed location, say x = x0. Then, the argument of the sine function in Equation (2) is constant -ωt. The displacement y, at fixed location, thus, varies sinusoidally with time. That is, the constituents of the medium at different positions execute simple harmonic motion. Finally, as t increases, x must increase in positive direction to keep kx- ωt + Φ constant. Thus, Eq. (2) represents a sinusoidal (harmonic) wave travelling along the positive direction of the x-axis. On the other hand, a function

y(x,t) = a sin(kx + ωt + Φ) ………………………………………………………(4)

represents a wave travelling in the negative direction of x-axis. Fig. (5) gives the names of the various physical quantities appearing in Eq. (2) that we now interpret.

y(x, t) : displacement as a function of position x and time t

a : amplitude of a wave

ω : angular frequency of the wave

k : angular wave number

kx – ωt + Φ : Initial phase angle (a +x = 0, t = 0)

Fig. 5 The meaning of standard symbols in Eq. (2)

Fig. 6 shows the plots of Eq. (2) for different values of time differing by equal intervals of time. In wave, the crest is the point of maximum positive displacement, the trough is the point of maximum negative displacement. To see how a wave travels, we can fix attention on a crest and see how it progresses with time. In the figure, this shown by a cross (×) on the crest. In the same manner, we can see the motion of a particular constituent of the medium at a fixed location, say at the origin of the x-axis. This is shown by a solid dot (♦). The plots of Fig. (6) show that with time, the solid dot (♦) at the origin moves periodically, i.e., the particle at the origin oscillates about its mean position as the wave progresses. This is true for any other location also. We also see that during the time the solid dot (♦) has completed one full oscillation, the crest has moved further by a certain distance. Fig. 6 – A harmonic wave progressing along the positive direction of x-axis at different times.he meaning of standard symbols in Eq. (2)

Using the plots of Fig. 6, we now define the various quantities of Eq. (2).

#### Amplitude and Phase

In Eq. (2), since the sine function varies between 1 and -1, the displacement y(x,t) varies between a and –a. We can take a to be a positive constant, without any loss of generality. Then, a represents the maximum displacement of the constituents of the medium from their equilibrium position. Note that the displacement y may be positive or negative, but a is positive. It is called the amplitude of the wave.

The quantity (kx- ωt + Φ) appearing as the argument of the sine function in Eq. (2) is called the phase of the wave. Given the amplitude a, the phase determines the displacement of the wave at any position and at any instant. Clearly Φ is the phase at x = 0 and t = 0. Hence, Φ is called the initial phase angle. By suitable choice of origin on the x-axis and the initial time, it is possible to have Φ=0. Thus there is no loss of generality in dropping Φ , i.e., in taking Eq. (2) with Φ= 0.

#### Wavelength and Angular Wave Number

The minimum distance between two points having the same phase is called the wavelength of the wave, usually denoted by λ. For simplicity, we can choose points of the same phase to be crests ot troughs. The wavelength is then the distance two consecutive crests or troughs in a wave. Taking Φ = 0 in Eq. (2), the displacement at t = 0 is given by

y(x,0) = a sin kx …………………………………………………(5)

Since the sine function repeats its value after every 2π change in angle,

sin kx = sin (kx + 2nπ) = sink(x + 2nπ/k)

That is the displacement at points x and at

x + 2nπ/k

are the same, where n = 1, 2, 3,…….. The least distance between points with the same displacement (at any given instant of time) is obtained by taking n=1. λ is then given by

λ = 2π/k or k = 2π/λ ………………………………….(6)

k is the angular wave number or propagation constant; its SI unit is radian per metre or rad m-1*

#### Period, Angular Frequency and Frequency

Fig. 7 shows again a sinusoidal plot. It describes not the shape of the wave at a certain instant but the displacement of an element (at any fixed location) of the medium as a function of time. We may for, simplicity, take Eq. (2) with Φ = 0 and monitor the motion of the element say at x = 0. We then get

y(0,t)= a sin(-ωt)

= –a sin-ωt

* Here again ‘radian’ could be dropped and the units could be written merely as m-1 . Thus, k represents 2π times the number of waves (or the total phase difference) that can be accommodated per unit length , with SI units m-1. Fig. 7 – An element of a string at a fixed location oscillates in time with amplitude a and period T, as the wave passes over it.

Now, the period of oscillation of the wave is the time takes for an element to complete on full oscillation. That is

a sin ωt = -a sin ω(t + T)

= –a sin(ωt + ωT)

Since sine function repeats after every 2π,

ωt = 2π or ω = 2π/T …………………………….(7)

ω is called the angular frequency of the wave. Its SI unit is rad s-1. The frequency γ is the number of oscillations per second. Therefore,

γ= 1/T = ω/2π ………………………………….(8)

γ is usually measured in hertz.

In the discussion above, reference has always been made to a wave travelling along a string or a transverse wave. In a longitudinal wave, the displacement of an element of the medium is parallel to the direction of propagation of the wave. In Eq. (2), the displacement function for a longitudinal wave is written as,

s(x, t) = a sin (kx – ωt + Φ) ………………………..(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. In Eq. (9), a is the displacement amplitude; other quantities 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).

Example 2
A wave travelling along a string is described by,

y(x, t) = 0.005 sin (8.0 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 wave length, 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?

On comparing this displacement equation with (2),

y(x, t) = a sin (k x – ω t),

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 m-1 and ω = 3.0 s-1

We, then, relate the wavelength λ to k through Eq. (6),

λ = 2π/k

= 2π/80.0 m-1

= 7.85 cm

(c) Now, we relate T to ω by the relation

T = 2π/ω

= 2π/3.0 s-1

= 2.09 s

and frequency, ν = 1/T = 0.48 Hz

The displacement y at x = 30.0 cm and time t = 20 s is given by

y = (0.005 m) sin (80.0×0.3 – 3.0×20)

= (0.005 m) sin (-36 + 12π)

= (0.005 m) sin (1.699)

= (0.005 m) sin (97°) ≅ 5 mm

### The Speed of a Travelling Wave

To determine the speed of propagation of a travelling wave, we can fix our attention on any particular point on the wave (characterised by some value of the phase) and see how that point moves in time. It is convenient to look at motion of the crest of the wave. Fig. 8 gives the shape of the wave at two instants of time, which differ by a small time interval Δt. The entire wave pattern is seen to shift to the right (positive direction of x-axis) by a distance Δx. In particular, the crest shown by a dot (♦) moves a distance Δx in time Δt. The speed of the wave is then Δx /Δt. Fig. 8 –
Progression of harmonic wave from time t to t +
Δt, where Δt is a small interval. The wave pattern as a whole shifts to the right. The crest of the wave (or a point with any fixed phase) moves right by the distance Δx in time Δt.

We can put the dot (♦) on a point with any other phase. It will move with the same speed v (otherwise the wave pattern will not remain fixed). The motion of a fixed phase point on the wave is given by

kx – ωt = constant ……………………………………….(10)

Thus,

kx – ωt = k(x+Δx) -ω(t + Δt)

or k Δx – ωΔt = 0

Taking Δx, Δt vanishingly small, this gives

dx/dt = ω/k = v …………………………………………….(11)

Relating ω to T and k to λ, we get

v = 2πν/2πλ = λν = λ/T …………………………….(12)

Eq. (12), a general relation for all progressive waves, shows that in the time required for one full oscillation by any constituent of the medium, the wave pattern travels a distance equal to the wavelength of the wave. It should be noted that the speed of a mechanical wave is determined by the inertial(linear mass density for strings, mass density in general) and elastic properties (Young’s modulus for linear media/shear modular, bulk modulus) of the medium. The medium determines the speed; Eq. (12) then relates wavelength to frequency for the given speed. Of course, as remarked earlier, the medium can support both transverse and longitudinal waves, which will have different speeds in the same medium. Later in this chapter, we shall obtain specific expressions for the speed of mechanical waves in some media.

#### Speed of a Transverse Wave on Stretched String

The speed of mechanical wave is determined by the restoring force set up in the medium when it is disturbed and the inertial properties (mass density) of the medium. The speed is expected to be directly related to the former and inversely to the latter. For waves on a string, the restoring force is provided by the tension t in the string. The inertial property will in this case be linear mass density µ, which is mass m of the string divided by its length L. Using Newton’s Laws of Motion, an exact formula for the wave speed on a string can be derived, but this derivation is outside the scope of this Chapter. We shall, therefore, use dimensional analysis. We already know that dimensional alone can never yield the exact formula. The overall dimensionless constant is always left undetermined by dimensional analysis.

The dimension of µ is [ML-1] and that of T is like force, namely [MLT-2]. We need to combine these dimensions to get the dimension of speed v [LT -1]. Simple inspection shows that the quantity T/µ has the relevant dimension

[MLT-2]/[ML-1] = [L²T-2]

Thus if T and µ are assumed to be the only relevant physical quantities,

v = C √(T/µ) ………………………………………………………(13)

where C is the undetermined constant of dimensional analysis. In the exact formula, it turns out C=1. The speed of transverse waves on a stretched string is given by

v = √(T/µ) ………………………………………………………(14)

Not the important point that the speed v depends only on the properties of the medium T and µ (T is a property of the stretched string arising due to an external force). It does not depend on wavelength or frequency of the wave itself. In higher studies, you will come across waves whose speed is not independent of frequency of the wave. Of the two parameters λ and ν the source of disturbance determines the frequency of the wave generated. Given the speed of the wave in the medium and

Propagation of a pulse on a rope You can easily see the motion of a pulse on a rope. You can also see its reflection from a rigid boundary and measure its velocity of travel. You will need a rope of diameter 1 to 3 cm, two hooks and some weights. You can perform this experiment in your classroom or laboratory.

Take a long rope or thick string of diameter 1 to 3 cm, and tie it to hooks on opposite walls in a hall or laboratory. Let one end pass on a hook and hang some weight (about 1 to 5 kg) to it. The walls may be about 3 to 5 m apart.

Take a stick or a rod and strike the rope hard at a point near one end. This creates a pulse on the rope which now travels on it. You can check the phase relation between the incident pulse and reflected pulse. You can easily watch two or three reflections before the pulse dies out. You can take a stopwatch and find the time for the pulse to travel the distance between the walls, and thus measure its velocity. Compare it with that obtained from Eq.(14)

This is also what happens with a thin metallic string of a musical instrument. The major difference is that the velocity on a string is fairly high because of low mass per unit length, as compared to that on a on a thick rope. The low velocity on a rope allows to watch the motion and make measurements beautifully.

frequency Eq. (12) then fixes the wave length

λ = v/ν …………………………………………………(15)

Example 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?

Mass per unit length of the wire,

µ = 5.0 × 10-3 kg/0.72 m

= 6.9 × 10-3 kg m-1

Tension T = 60 N

The speed of wave on the wire is given by

v = √(T/µ) = √(60 N/6.9 × 10-3kg m-1) = 93 m s-1

#### Speed of a Longitudinal Wave (Speed of Sound)

In a longitudinal wave, the constituents of the medium oscillate forward and backward in the direction of propagation of the wave. We have already seen that the sound waves travel in the form of compressions and rarefactions of small volume elements of air. The elastic property that determines the stress under compressional strain is the bulk modulus of the medium defined by ‘Mechanical Properties of Solids’.

b = -ΔP/(ΔV/V) …………………………………………….(16)

Here, the change in pressure ΔP produces a volumetric strain ΔV/V. B has the same dimension as pressure and given in SI units in terms of pascal (Pa). The inertial property relevant for the propagation of wave is the mass density ρ, with dimensions [ML-3]. Simple inspection reveals that quantity B/ρ has the relevant dimension:

[ML-1T-2]/[ML-3] = [L2T-2] ………………………..(17)

Thus, if B and ρ are considered to be the only relevant physical quantities,

v = C√(B/ρ) ……………………………………………(18)

where, as before, C is the fundamental constant from dimensional analysis. The exact derivation shows that C=1. Thus, the general formula for waves in a medium is:

v = C√(B/ρ) ……………………………………………(19)

For a linear medium, like a solid bar, the lateral expansion of the bar is negligible and we may consider it to be only under longitudinal strain. In that case, the relevant modulus of elasticity is Young’s modulus, which has the same dimension as the Bulk modulus. Dimensional analysis for this case is the same as before and yields a relation like Eq. (18), with an undetermined C, which the exact derivation shows to be unity. Thus, the speed of longitudinal waves in a solid bar is given by

v = √(Y/ρ) ……………………………………………(20)

where Y is the Young’s modulus of the material of the bar. Table 1 gives the speed of sound in some media.

Medium Speed (m s-1)
Gases
Air (0°C) 331
Air (20°C) 331
Helium 965
Hydrogen 1284
Liquids
Water (0°C) 1402
Water (20°C) 1482
Seawater 1522
Solids
Aluminium 6420
Copper 3560
Steel 5941
Granite 6000
Vulcanised Rubber 54

Liquids and solids generally have higher speed of sound than gases. [Note for solids, the speed being referred to is the speed of longitudinal waves in the solid]. This happens because they are much more difficult to compress than gases and so have much higher values of bulk modulus. Now, see Eq. (19). Solids and liquids have higher mass densities (ρ) than gases. But the corresponding increase in both modulus (B) of solids and liquids is much higher. This is the reason why the sound waves travel faster in solids and liquids.

We can estimate the speed of sound in a gas in the ideal gas approximation. For an ideal gas, the pressure P, volume V and temperature T are related by (see Chapter ‘Thermal Properties of Matter).

PV = NkBT …………………………………………………….(21)

where N is the number of molecules in volume V, kB is the Boltzmann constant and T the temperature of the gas (in Kelvin). Therefore, an isothermal change it follows from Eq. (21) that

P + PΔV = 0

or –ΔP/(ΔV/V) = P

Hence, substituting in Eq. (16) the speed of a longitudinal wave in an ideal gas is given by,

v = √(P/ρ) ………………………………………………(22)

This relation was first given by Newton and is known as Newton’s Formula.

Example 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.

We know that 1 mole of any gas occupies 22.4 litres at STP. Therefore, density of air at STP is:

ρ0 = (mass of one mole of air)/(volume of one of air at STP)

= 29.0 × 10-3 kg/29.0 × 10-3 m3

= 1.29 kg m-3

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

v = [(1.01 × 105 N m-2)/1.29 kg m-3] = 280 m s-1 ……………..(23)

The result shown in Eq. (23) is about 15% smaller as compared to the experimental value of 331 m s-1 as given in Table 1. Where did we wrong? 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. 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.

PVγ = constant

i.e., Δ(PVγ)= 0

or PγVγ-1ΔV + VγΔP = 0

where γ is the ratio of two specific heats, Cp/Cv.

Thus, for an ideal gas the adiabatic bulk modulus is given by,

= γP

The speed of sound is, therefore, from Eq. (19), given by,

v = √(γP/ρ) ………………………………………………….(24)

This modification of Newton’s formula is referred to as the Laplace correction. For air γ = 7/5. Now using Eq. (24) 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.

### The Principle of Superposition of Waves

What happens when two wave pulses travelling in opposite directions cross each other Fig. (9)? It turns out that wave pulses continue to retain their identities after they have crossed. However, during the time they overlap, the wave pattern is different from either of the pulses of equal and opposite shapes move towards each other. When the pulses overlap, the resultant displacement due to each pulse. This is known as the principle of superposition of waves. Fig. 9 – Two pulses having equal and opposite displacements moving in opposite directions. The overlapping pulses add up to zero displacement in curve (c).

According to this principle, each pulse moves as if others are not present. The constituents of the medium, therefore, suffer displacements due to both and since the displacements can be positive and negative, the net displacement is an algebraic sum of the two. Fig. 9 gives graphs of the wave shape at different times. Note the dramatic effect in the graph (c); the displacements due to the two pulses have exactly cancelled each other and there is zero displacement throughout.

To put the principle of superposition mathematically, let y1(x,t) and y2(x,t) be the displacements due to two wave disturbances in the medium. If the waves arrive in a region simultaneously, and therefore, overlap, the net displacement y(x,t) is given by

y(x,t) = y1(x,t)+ y2(x,t) ………………………………………….(25)

If we have two or more waves moving in the medium the resultant waveform is the sum f wave functions of individual waves. That is, if the wave functions of the moving waves are

y1= f1(x – vt)

y2= f2(x – vt)

………………..

………………..

yn= fn(x – vt)

then the wave function describing the disturbance in the medium is

y = f1(x – vt) + f2(x – vt) + ………….+ fn(x – vt)

=fi(x – vt) The principle of superposition is basic to the phenomenon is basic to the phenomenon of interference.

For simplicity, consider two harmonic travelling waves on a stretched string, both with the same ω (angular frequency) and k (wave number) and, therefore, the same wavelength λ. Their wave speed will be identical. Let us further assume that their amplitudes are equal and they are both travelling in the positive direction of x-axis. The waves only differ in their initial phase. According to Eq. (2) the two waves are described by the functions:

y1(x,t)= a sin (kx-ωt) …………………………………………..(27)

y2(x,t)= a sin (kx-ωt + Φ) …………………………………………..(28)

The net displacement is then, by the principle of superposition, given by

y(x,t)= a sin (kx-ωt) + a sin (kx-ωt + Φ) …………………………..(29)

= a[2 sin[{(kx-ωt) + (kx-ωt + Φ}/2]cosΦ/2] …………..(30)

where we have used the familiar trigonometric identity for sin A + sin B. We then have

y(x,t) = 2 a cosΦ/2 sin (kx-ωt + Φ) …………………………..(31)

Eq. (31) is also a harmonic travelling wave in the positive direction of x axis, with the same frequency and wave length. However, its initial phase angle is Φ/2. The significant thing is that its amplitude is a function of the phase difference Φ between the constituent two waves:

A(Φ) = 2 a cos ½Φ ………………………………………….(32)

For Φ = 0, when the waves are in phase,

y(x,t)= 2a sin (kx-ωt) …………………………..(33)

i.e., the resultant wave has amplitude 2a, the largest possible value for A. For For Φ = π, the waves are completely, out of phase and the resultant wave has zero Fig 10 – The resultant of two harmonic waves of equal amplitude and wavelength according to the principle of superposition. The amplitude of the resistant wave depends on the phase difference Φ, which is zero for (a) and π for (b)

displacement everywhere at all times

y (x,t)= 0 …………………………………………………………………..(34)

Eq. (33) refers to the so-called constructive interference of the two waves where the amplitudes add up in the resultant wave. Eq. (34) is the case of destructive interference where the amplitudes subtract out in the resultant wave. Fig. 10 shows these two cases of interference where the amplitudes subtract out in the resultant wave. Fig. 10 shows these two cases of interference of waves arising from the principle of superposition.

### Reflection of Waves

So far we considered waves propagating in an unbounded medium. What happens if a pulse or a wave meets a boundary? If the boundary is rigid, the pulse or wave gets reflected.The phenomenon of echo is an example of reflection by a rigid boundary. If the boundary is not completely rigid or is an interface between two different elastic media, the situation is some what complicated. A part is transmitted in to the second medium. If a wave is incident obliquely on the boundary between two different media the transmitted wave is called the refracted wave. The incident and refracted waves obey Snell’s law of refraction, and the incident and reflected waves obey the usual laws of reflection.

Fig. 11 shows a pulse travelling along a stretched string and being reflected by the boundary. Assuming there is no absorption of energy by the boundary, the reflected wave has the same shape as the incident pulse but it suffers a phase change of π or 180° on reflection. This is because the boundary is rigid and the disturbance must have zero displacement at all times at the boundary. By the principle of superposition, this is possible only if the reflected and incident waves differ by a phase of π, so that the resultant displacement is zero. This reasoning is based on boundary condition on a rigid wall. We can arrive at the same conclusion dynamically also. As the pulse arrives at the wall, it exerts a force on the wall. By Newton’s Third Law, the wall exerts an equal and opposite force on the string generating a reflected pulse that differs by a phase of π. Fig 11 – Reflection of a pulse meeting a rigid boundary

If on the other hand, their boundary point is not rigid but completely free to move (such as in the case of a string tied to a freely moving ring on a rod), the reflected pulse has the same phase and amplitude (assuming no energy dissipation) as the incident pulse. The net maximum displacement at the boundary is the open end of an organ pipe.

To summarise, a travelling wave or pulse change of π on reflection at an open boundary. To put this mathematically, let the incident travelling wave be

y2(x,t)= a sin (kx-ωt)

At a rigid boundary, the reflected wave is given by

yr(x,t)= a sin (kx-ωt + π)

= –a sin (kx-ωt) ……………………………………………….(35)

At an open boundary, the reflected wave is given by

yr(x,t)= a sin (kx-ωt + 0)

= a sin (kx-ωt) ……………………………………………….(36)

Clearly, at the rigid boundary, y = y2 + yr = 0 at all times.

#### Standing Waves and Normal Modes

We considered above reflection at one boundary. But there are familiar situations (a string fixed at either end or an air column in a pipe with either end closed) in which reflection takes place at two or more boundaries. In a string, for example, a wave travelling in one direction will get reflected from the other end. This will go on until there is a steady wave pattern set up on the string. Such wave patterns are called standing waves or stationary waves. To see this mathematically, consider a wave travelling along the positive direction of x-axis and a reflected wave of the same amplitude and wavelength in the direction of x-axis. From Eqs. (2) and (4), with Φ = 0, we get:

y1(x,t)= a sin (kx-ωt)

y2(x,t)= a sin (kx+ωt)

The resultant wave on the string is, according to the principle of superposition:

y(x,t)= y1(x,t)+ y1(x,t)

= a[sin(kx-ωt) + sin (kx+ωt)]

Using the familiar trigonometric identity

Sin (A+B) + Sin (A-B) = 2 sin A cos B we get,

y(x,t)= 2 a sin kx cos ωt ………………………………………(37)

Note the important difference in the wave pattern described by Eq. (37) from that described by Eq. (2) or Eq. (4). The terms kx and ωt appear separately, not in the combination of kx-ωt. The amplitude of this wave is 2 a sin kx. Thus, in this wave pattern, the amplitude varies from point to point, but each element of the string oscillates with the same angular frequency ω or time period. There is no phase difference between oscillations of different elements of the wave. The string as a whole vibrates in phase with differing amplitudes at different points. The wave pattern is neither moving to the right nor to the left. Hence, they are called standing or stationary waves. The amplitude is fixed at a given location but, as remarked earlier it is different at different locations. The points at which the amplitude is zero (i.e. where there is no motion at all) are called nodes; the points at which the amplitude is the largest are called antinodes. Fig. 12 shows a stationary wave pattern resulting from a superposition of two travelling waves in opposite directions.

The most significant feature of stationary waves is that the boundary conditions constrain the possible wavelengths or frequencies of vibration of the system. The system cannot oscillate with any arbitrary frequency (contrast this with a harmonic travelling wave), but is characterised by a set of natural frequencies or normal modes of oscillation. Let u determine these normal modes for a stretched string fixed at both ends.

First, from Eq. (37), the positions of nodes (where the amplitude is zero) are given by

sin kx = 0,

which implies

kx = nπ; n = 0, 1, 2, 3, ………

Since, k = nλ/2; we get

x = nλ/2; n = 0, 1, 2, 3, ……… …………………….(38)

Clearly, the distance between any two successive nodes is λ/2. In the same way, the positions of antinodes (where the amplitude is the largest) are given by the the largest value of sin kx: Fig 12 – 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.

I sin kx I = 1

which implies

kx = [n + ½] λ/2; n = 0, 1, 2, 3, ……… ………………………(39)

Again the distance between any two consecutive antinodes is λ/2. Eq.  can be applied to thee case of a stretched string of length L fixed at both ends. Taking one end to be at x=0 and x = L are positions of nodes. The x = 0 condition is already satisfied. The x = L node condition requires that the length L is related to λ by

L = n λ/2; n = 1, 2, 3, ……… ………………………(40)

Thus, the possible wavelengths of stationary waves are constrained by the relation

λ = 2 L/n; n = 1, 2, 3, ……… ………………………(41)

with corresponding frequencies

ν = n v/2 L, for n = 1, 2, 3, ……… ………………………(42)

We have thus obtained the natural frequencies -the normal modes of oscillation of the system. The lowest possible natural frequency of a system is called its fundamental mode or the first harmonic. For the stretched string fixed at either end it is given by ν = v/2L, corresponding to n = 1 of Eq. (42). Here ν is the speed of wave determined by the properties of the medium. The n=2 frequency is called the second harmonic; n= 3 is the third harmonic and so on. We can label the various harmonics by the symbol νn(n=1, 2, ……).

Fig. 13 shows the first six harmonics of a stretched string fixed at either end. A string need not vibrate in one of these modes only. Generally, the vibration of a string will be a superposition of different modes; some modes may be more strongly excited and some less. Musical instruments like sitar or violin are based on this principle. Where the string is plucked or bowed, determines which modes are prominent than others.

Let us next consider normal modes of oscillation of an air column with one end closed and the other open. A glass tube partially filled with water illustrates this system. The end in contact with water is a node, while the open end is an antinode. At the node the pressure changes are the largest, while the displacement is minimum (zero). At the open end – the antinode, it is just the other way- least pressure change and maximum amplitude of displacement. Taking the end in contact with water to be x=0, the node condition (Eq. 38) is already satisfied. If the other end x = L is an antinode, Eq. (39) gives

L = (n+½)λ/2, for n= 0, 1, 2, 3, …..

The possible wavelengths are then restricted by the relation:

λ = 2 L/(n+½) , for n= 0, 1, 2, 3, ….. ………………………(43)

The normal modes – natural frequencies – of the system are

ν = (n+½)v/2L; n= 0, 1, 2, 3, ….. ………………………(44)

The fundamental frequency corresponds to n = 0, and is given by v/4L. Fig 13 – The first six harmonics of vibrations of a stretched string fixed at both ends.

The higher frequencies are odd harmonics. i.e., odd multiples of the fundamental frequency : 3v/4L, 5v/4L, etc. Fig. 14 shows the first six odd harmonics of air column with one end closed and the other open. For a pipe open at both ends, each end is an antinode. It is then easily seen that an open air column at both ends generates all harmonics (see Fig. 15).

The systems above, strings and air columns, can also undergo forced oscillations. If the external frequency is close to one of the natural frequencies, the system shows resonance.

Normal modes of a circular membrane rigidly clamped to the circumference as in a tabla are determined by the boundary condition that no point on the circumference of the membrane vibrates. Estimation of the frequencies of normal modes of this system is more complex. This problem involves wave propagation in two dimensions. However, the physics is the same.

Example 5
A pipe, 30.0 cm long, is open at both ends. Which harmonic mode of the pipe resonates 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.

The first harmonic frequency is given by

ν1 = v/λ1 = v/2L (open pipe)

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

νn =nv/2L, for n = 1, 2, 3, ……….(open pipe)

First few modes of an open pipe are shown in Fig. 15  Fig 14 – Normal modes of an air column open at one end closed at the other end. Only the odd harmonics are seen to be possible.

For L = 30.0 cm, v = 330 m s-1,

νn =n 330 m s-1/0.6 (m) = 550 n s-1

Clearly, a source of frequency 1.1 kHz will resonate at v2. i.e., the second harmonic.

Now if one end of the pipe is closed (Fig 15) it follows from Eq. (50) that the fundamental frequency is

ν1 = v/λ1 = v/4L (pipe closed at one end)

and only the odd numbered harmonics are present:

ν3 = 3v/4L, ν5= 5v/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 forth harmonic. Since this harmonic is not a possible mode, no resonance will be observed with the source, the moment one end is closed.

### Beats

‘Beats’ is an interesting phenomenon arising from interference of waves. When two harmonic sound waves of close (but not equal) frequencies are heard at the same time, we hear a sound of similar frequency (the Fig. 15 – Standing waves in an open pipe, first four harmonics depicted.

average of two close frequencies), but we hear something else also. We hear audibly distinct waxing and waning of the intensity of the sound, with a frequency equal to the difference in the two close frequencies. Artists use this phenomenon often while tuning their instruments with each other. They go on tuning until their sensitive ears do not detect any beats.

To see this mathematically, let us consider two harmonic sound waves of nearly equal angular frequency ω1 and ω2 and fix the location to be x = 0 for convenience. Eq. (2) with a suitable choice of phase (Φ = π/2 for each) and assuming equal amplitudes, gives

s1 = a cos ω1t and s2 = a cos ω2t ……………………………………….(45)

Here we have placed the symbol y by s, since we are referring to longitudinal not transverse displacement. Let ω1 be the (slightly) greater of the two frequencies. The resultant displacement is, by principle of superposition,

s = s1 + s2 = a (cos ω1t +cos ω2t)

Using the familiar trigonometric identity for cos A + cos B, we get

= 2 a cos[1– ω2)t/2] cos[1+ ω2)t/2] ………………………..(46)

which may be written as:

s = [2 a cos ωbt]cos ωat ……………………………………(47)

If I ω1– ω2 I « ω1, ω2, ωa » ωb, where

ωb = (ω1– ω2)/2 and ωa = (ω1+ ω2)/2

Now if we assume I ω1– ω2 I « ω1 which means ωa » ωb we can interpret

Musical Pillars Temples often have some pillars portraying human figures playing musical instruments, but seldom do these pillars themselves produce music. At the Nellaiappar temple in Tamil Nadu, gentle taps on a cluster of pillars carved out of a single piece of rock produce the basic notes of Indian classical music, viz. Sa, Re, Ga, Ma, Pa, Dha, Ni, Sa. Vibration of these pillars depend on elasticity of the stone used, its density and shape.

Musical pillars are categorised in to three types: The first is called the Shruti Pillar, as it can produce the basic notes – the “swaras”. The second type is the Gana Thoongal, which generates the basic tunes that make up the “ragas”. The third variety is the Laya Thoongal pillars that produce “taal” (beats) when tapped. These pillars at Nellaiappar temple are a combination of the Shruti and Laya types.

Archaeologists date the Nelliappar temple to the 7th century and claim it was built by successive rulers of the Pandyan dynasty.

The musical pillars of Nelliappar and several other temples in southern India like those at Hampi (picture), Kanyakumari, and Thiruvananthapuram are unique to the country and have no parallel in any other part of the world.

Eq. (47) as follows. The resultant wave is oscillating with the average angular frequency ωa; however its amplitude is not constant in time, unlike a pure harmonic wave. The amplitude is the largest when the term cos ωbt takes its limit +1 or -1. In other words, the intensity of the resultant wave waxes and wanes with a frequency which is 2ωb = ω1ω2. Since ω = 2πν, the beat frequency νbeat, is given by

νbeat = ν1 ν2 …………………………………………….(48)

Fig. 16 illustrates the phenomenon of beats for two harmonic waves of frequencies 11 Hz and 9 Hz. The amplitude of the resultant wave shows beats at a frequency of 2 Hz. Fig 16 – Superposition of two harmonic waves, one of frequency 11 Hz (a), and the other of frequency 9 Hz (b), giving rise to beats of frequency 2 Hz, as shown in (c).

Example 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?

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

### Doppler Effect

It is an everyday experience that the pitch (or frequency) of the whistle of a fast moving train decreases as it reaches away. When we approach a stationary source

Reflection of Sound in an open pipe When a high pressure pulse of air travelling down an open pipe reaches the other end, its momentum drags the air out in to the open, where pressure falls rapidly to the atmospheric pressure. As a result the air following after it in the tube is pushed out. The low pressure at the end of tube draws air from further up the tube. The air gets drawn towards the open end forcing the low pressure region to move upwards. As a result a pulse of high pressure air travelling down the tube turns in to a pulse of low pressure air travelling up tube. We say a pressure wave has been reflected at the open end with a change in phase of 180°. Standing waves in an open pipe organ like the flute is a result of this phenomenon.

Compare this with what happens when a pulse of high pressure air arrives at a closed end: it collides and a result pushes the air back in the opposite direction. Here, we say that the pressure wave is reflected, with no change in phase.

of sound with high speed, the pitch of the sound heard appears to be higher than that of the source. As the observer recedes away from the source, the observer recedes away from the source, the observed pitch (or frequency) becomes lower than that of the source. This motion- related frequency change is called Doppler effect. The Austrian physicist Johann Christian Doppler first proposed the effect in 1842. Buys Ballot in Holland tested it experimentally in 1845. Doppler effect is a wave phenomenon, it holds not only sound waves but also for electromagnetic waves. However, here we shall consider only sound waves.

We shall analyse changes in frequency under three different situations: (1) observer is stationary but the source is moving, (2) observer is moving but the source is stationary, and (3) both the observer and the source are moving. The sources (1) and (2) differ from each other because of the absence or presence of relative motion between the observer and the medium. Most waves require a medium for their propagation; however, electromagnetic waves do not require any medium for propagation. If there is medium present, the Doppler shifts are same irrespective of whether the source moves or the observer moves, since there is no way of distinction between the two situations.

#### Source Moving: Observer Stationary

Let us choose the convention to take the direction from the observer to the source as the positive direction of velocity. Consider a source S moving with velocity vs and an observer who is stationary in a frame in which the medium is also at rest. Let the speed of a wave of angular frequency ω and period T0, Both measured by an observer at rest with respect to the medium, be v. We assume that the observer has a detector that counts every time a wave crest reaches it. As shown in Fig. 17, at time t=0 the source is at point S1 located at a distance L from the observer at time t1 = L/v. At time t = T0 the source has moved a distance (L + vsT0) from the observer. At S2, located at a distance (L + vsT0) from the observer. At S2, the source emits a second crest. This reaches the observer at Fig 17 – Doppler effect (change in frequency of wave) detected when the source is moving and the observer is at rest in the medium.

t2 = T0 + (L + vsT0)/v

At time nT0, the sorce emits its (n+1)th crest and this reaches the observer at time

tn+1 = nT0 + (L + nvsT0)/v

Hence, in a time interval

[nT0 + [L + nvsT0 -L/v]

the observer’s detector counts n crests and the observer record the period of the wave as T given by

T = [nT0 + [L + nvsT0 -L/v]/n

= T0 + vsT0/v

= T0 (1+ vs/v) …………………………………………(49)

Equation (49) may be rewritten in terms of the frequency ν0 that would be measured if the source and observer were stationary, and the frequency ν observed when the source is moving, as

ν = ν0 (1+ vs/v)-1 …………………………………………..(50)

If vs is small compared wit the wave speed v, taking binomial expansion to terms in first order in vs/v and neglecting higher power, Eq. (50) may be approximated, giving

ν = ν0 (1 – vs/v) …………………………………………..(51)

For a source approaching the observer, we replace vs by –vs to get

ν = ν0 (1 + vs/v) …………………………………………..(52)

The observer thus measures a lower frequency when the source recedes from him than he does when it is at rest. He measures a higher frequency when the source approaches him.

#### Observer Moving: Source Stationary

Now to derive the Doppler shift when the observer is moving with velocity v0 towards the source and the source is at rest, we have to proceed in a different manner. We work in the reference frame of the moving observer. In this reference frame the source and medium are approaching at speed v0 and the speed with which the wave approaches is v0+ v. Following a similar procedure as in the previous case, we find that the time interval between the arrival of the first and the (n+1)th crest is

tn+1 – t1= nT0 [nvsT0/(v0+ v)]

The observer thus, measures the period of the wave to be

= T0 [1-(v0/v0 + v)]

= T0 [1+v0/v]-1

giving

ν = ν0 [1+v0/v] …………………………………………(53)

If v0/v is small, Doppler shift is almost same whether it is observer or the source moving since Eq. (53) and the approximate relation Eq. (51) are the same.

#### Both Source and Observer Moving

We will now derive a general expression for Doppler shift when both the source and the observer are moving. As before, let us take the direction from the observer to the source as the positive direction. Let the source and the observer be moving with velocities vs and v0 respectively as shown in Fig. 18. Suppose at time t = 0, the observer is at O1 and the source is at S1. O1 being to the left of S1. The source emits a wave of velocity v, of frequency v and period T0 all measured by an observer at rest with respect to the medium. Let L be the distance between O1 and S1 at t = 0, when the source emits the first crest. Now, since the observer is moving, the velocity of the wave relative to the observer is v + v0. At time t = T0, both the observer and the source have moved to their new positions O2 and S2 respectively. The new distance between the observer and the source, O2 S2 would be L + (vs-v0)T0. At S2, the source emits a second crest . Fig 18 – Doppler effect when both the source and observer are moving with different velocities.

Application of Doppler effect

The change in frequency caused by a moving object due to Doppler effect is used to measure their velocities in diverse areas such as military, medical science, astrophysics, etc. It is also used by police to check over-speeding of vehicles.

A sound wave or electromagnetic wave of known frequency is sent towards a moving object. Some part of the wave is reflected from the object and its frequency is detected by the monitoring station. This change in frequency is called Doppler shift.

It is used at airports to guide at 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 parts of the body. Here they use ultrasonic 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 the heart valves, as well as pulsation of the heart of the foetus. In the case of heart, the picture generated is called echo cardiogram.

This reaches the observer at time.

t2 = T0 + (L + (vs -v0)T0)/(v + v0)

At time n T0 the source emits its (n+1)th crest and this reaches the observer at time

tn+1 = nT0 + (L + n(vs -v0)T0)/(v + v0)

Hence, in a time interval tn+1 t1, i.e.,

nT0 + (L + n(vs -v0)T0)/(v + v0) – L/(v + v0),

the observer counts n crests and the observer records the period of the wave as equal to T given by

T = T0 + [(1 + (vs -v0)/(v + v0)] = T0(v + vs)/(v + v0) …………….(54)

The frequency ν observed by the observer is given by

ν = ν0 (v + v0)/(v + vs) ……………………………………(55)

Consider a passenger sitting in a train moving on a straight track. Suppose she hears a whistle sounded by the driver of the train. What frequency will she measure or hear? Here both the observer and the source are moving with the same velocity, so there will be no shift in frequency and the passenger will note the natural frequency. But an observer outside who is stationary with respect to the track will note a higher frequency if the train is approaching him and a lower frequency when it recedes from him.

Note that we defined the direction from the observer to the source as the positive direction. Therefore, if the observer is moving towards the source, v0 has a positive (numerical) value whereas if O is moving away from S, v0 has a negative value. On the other hand, if S is moving away from O, vs has a positive value whereas if it is moving towards O, vs has a negative value. The sound emitted by the source travels in all directions. It is that part of sound coming towards the observer receives and detects. Therefore, the relative velocity of sound with respect to the observer is v+ v0 in all cases.

Example 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.

(1) The observer is at rest and the source is moving with a speed of 200 ms-1, we must use Eq. (50) and not the approximate Eq. (51). Since the source is approaching a stationary target, v0 =0, and vs must be replaced by –vs. Thus, we have

ν = ν0 (v + vs)/v)-1,

ν = 1000 Hz × [1-200 ms-1/330 ms-1]

≈ 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 vs = 0 and v0 has a positive value. The frequency of the sound emitted by the source (target) is ν, the frequency intercepted by the target and not ν0. Therefore, the frequency as registered by the rocket is

ν’ = ν (v + v0)/v)

= 2540 Hz × (200 ms-1+330 ms-1)/330 ms-1

4080 Hz 