Physics THE SPEED OF A TRAVELLING WAVE ,TRANSVERSE WAVE & LONGITUDINAL WAVE AND REFLECTION OF WAVES

### Topic Covered

color{blue}{star} THE SPEED OF A TRAVELLING WAVE
color{blue}{star} SPEED OF A TRANSVERSE WAVE ON STRETCHED STRING
color{blue}{star} SPEED OF A LONGITUDINAL WAVE (SPEED OF SOUND)
color{blue}{star} REFLECTION OF WAVES

### THE SPEED OF A TRAVELLING WAVE

color{blue} ✍️ To determine the speed of propagation of a travelling wave, we can fix our attention on any particular point on the wave (characterized by some value of the phase) and see how that point moves in time.

color{blue} ✍️ It is convenient to look at the motion of the crest of the wave. Fig. 15.8 gives the shape of the wave at two instants of time which differ by a small time internal Δt.

color{blue} ✍️ 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 cross (×) moves a distance Δx in time Δt, The speed of the wave is then Δx//Δt.

color{blue} ✍️We can put the cross (×) 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"

...............................(15.10)

color{blue} ✍️Thus, as time t changes, the position x of the fixed phase point must change so that the phase remains constant. Thus

color {blue} {kx – ωt = k(x+Δx) – ω(t+Δt)}

or color{purple} {k Δx – ω Δt =0}

color{blue} ✍️Taking Δx, Δt vanishingly small, this gives

color{blue} {(dx)/(dt) = omega/k = v}

...................(15.11)

Relating ω to T and k to λ, we get

color{blue} {v = ( 2 n v)/ ( 2 pi k ) = lamda v = lamda/T}

......................(15.12)

color{blue} ✍️Eq. (15.12), 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.

color{blue} ✍️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 modulus, bulk modulus) of the medium.

color{blue} ✍️The medium determines the speed; Eq. (15.12) 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.

### Speed of a Transverse Wave on Stretched String

color{blue} ✍️The speed of a mechanical wave is determined by the restoring force setup in the medium when it is disturbed and the inertial properties (mass density) of the medium.

color{blue} ✍️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.

color{blue} ✍️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 book.

color{blue} ✍️We shall, therefore, use dimensional analysis. We already know that dimensional analysis alone can never yield the exact formula. The overall dimensionless constant is always left undetermined by dimensional analysis.

color{blue} ✍️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]) = [ L^2 T^-2]

color{blue} ✍️Thus if T and μ are assumed to be the only relevant physical quantities,

color{blue} {nu = C sqrt(T/mu)}

.......................(15.13)

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

color{blue} {nu = sqrt(T/mu)}

....................... (15.14)

color{blue} ✍️Note 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).

color{blue} ✍️It does not depend on wave length or frequency of the wave itself. In higher studies, you will come across waves whose speed is not independent of frequency of the wave.

color{blue} ✍️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 the frequency Eq. (15.12) then fixes the wavelength

color{blue} {lamda = nu/v}

................(15.15)
Q 3169278115

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 ?
Class 11 Chapter 15 Example 3
Solution:

Mass per unit length of the wire,

mu = (5.0 xx 10^-3 kg)/(0.72 m)

= 6.9 xx 10^-3  kg m^-1

Tension, T = 60 N

The speed of wave on the wire is given by

nu = sqrt (T/mu) = sqrt ((60 N) / ( 6.9 xx 10^-3 kg m^-1 ) = 93 ms^-1

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

color{blue} ✍️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.

color{blue} ✍️The elastic property that determines the stress under compressional strain is the bulk modulus of the medium defined by (see chapter 9)

color{blue} {B = - (Delta P)/ ( Delta V//V)

..................... (15.16)

color{blue} ✍️Here the change in pressure ΔP produces a volumetric strain (DeltaV)/V B has the same dimension as pressure and given in SI units in terms of pascal (Pa).

color{blue} ✍️The inertial property relevant for the propagation of wave in the mass density ρ, with dimensions [ML^-3]. Simple inspection reveals that quantity B//ρ has the relevant dimension:

color{blue} { ([ML^-1 T^-2] ) / ([ML^-3]) = [ L^2 T^-2]}

....................... (15.17)

color{blue} ✍️Thus if B and rho are considered to be the only relevant physical quantities,

color{blue} {nu = C sqrt (B/rho)}

......................... (15.19)

color{blue} ✍️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 in Young’s modulus, which has the same dimension as the Bulk modulus.

color{blue} ✍️Dimensional analysis for this case is the same as before and yields a relation like Eq. (15.18), 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

color{blue} {nu =sqrt(Y/rho)}

...........(15.20)

color{blue} ✍️where Y is the Young’s modulus of the material of the bar. Table 15.1 gives the speed of sound in some media.

color{blue} ✍️Liquids and solids generally have higher speeds of sound than in gases. [Note for solids, the speed being referred to is the speed of longitudinal waves in the solid].

color{blue} ✍️This happens because they are much more difficult to compress than gases and so have much higher values of bulk modulus. This factor more than compensates for their higher densities than gases.

color{blue} ✍️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 11).

color{blue} { PV = NkBT}

....................... (15.21)

color{blue} ✍️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).

color{blue} ✍️Therefore, for an isothermal change it follows from Eq.(15.21) that

color{blue}{VΔP + PΔV = 0}

or

color{blue} {- (Delta P)/(Delta V//V) = P}

color{blue} ✍️Hence, substituting in Eq. (15.16), we have

B = P

color{blue} ✍️Therefore, from Eq. (15.19) the speed of a longitudinal wave in an ideal gas is given by,

color{blue} {nu = sqrt (P/rho)}

.....................(15.22)

color{blue} ✍️This relation was first given by Newton and is known as Newton’s formula.

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

color{purple}{nu = [ ( 1.01 xx 10^5 N m^-2 )/ ( 1.29 kg m^-3) ] ^(1//2)}

color{blue} {nu= 280 ms^-1}

...................(15.23)

color{blue} ✍️The result shown in Eq.(15.23) is about 15% smaller as compared to the experimental value of 331 m s^-1 as given in Table 15.1.

color{blue} ✍️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.

color{blue} ✍️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 color{purple} {P^γ V ^(γ –1 ) ΔV + V^γ ΔP = 0}

color{blue} ✍️Thus, for an ideal gas the adiabatic bulk modulus is given by,

color{ blue } {B_(ad) = - (Delta P)/ ( Delta V//V)} = color{blue} {= gamma P}

where γ is the ratio of two specific heats, C_p//C_v. The speed of sound is, therefore, given by,

color{ blue} { nu = sqrt ( gamma P/ rho)

.............. (15.24)

color{blue} ✍️This modification of Newton’s formula is referred to as the Laplace correction. For air γ = 7//5. Now using Eq. (15.24) 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.
Q 3169478315

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.
Class 11 Chapter 15 Example 4
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)

= (29.0 xx 10^-3 kg )/( 22.4 xx 10^-3 m^3)

= 1.29 kg m^-3

### THE PRINCIPLE OF SUPERPOSITION OF WAVES

color{blue} ✍️If two wave pulses travelling in opposite directions cross each other, It turns out that wave pulses continue to retain their identities after they have crossed.

color{blue} ✍️However, during the time they overlap, the wave pattern is different from either of the pulses. Figure 15.9 shows the situation when two pulses of equal and opposite shapes move towards each other.

color{blue} ✍️When the pulses overlap, the resultant displacement is the algebraic sum of the displacement due to each pulse. This is known as the principle of superposition of waves. According to this principle, each pulse moves as if others are not present.

color{blue} ✍️The constituents of the medium therefore suffer displacement due to both and since displacements can be positive and negative, the net displacement is an algebraic sum of the two. Fig. 15.9 gives graphs of the wave shape at different times.

color{brown} {"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.

color{blue} ✍️To put the principle of superposition mathematically, let y_1 (x,t) and y_2 (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

color{ blue} {y (x, t) = y_1(x, t) + y_2(x, t)}

........................... (15.25)

color{blue} ✍️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)

color{blue} ✍️then the wave function describing the disturbance in the medium is

y = f_1(x – vt)+ f_2(x – vt)+ ...+ fn(x – vt)

color{blue } {= sum_(i =1)^n f_i ( x - v t)}

........................ (15.26)

color{blue} ✍️The principle of superposition is basic to the phenomenon of interference.

color{blue} ✍️For simplicity, consider two harmonic travelling waves on a stretched string, both with the same ω (angular frequens) and k (wave number), and, therefore, the same wavelength λ.

color{blue} ✍️Their wave speeds 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. (15.2), the two waves are described by the functions:

color{blue } { y_1(x, t) = a sin (kx – ωt)}

..........................(15.27)

and

color{blue} {y_2(x, t) = a sin (kx – ωt + φ )}

.....................(15.28)

color{blue} "✍️ The net displacement is then, by the principle of superposition" , given by

color{blue} {y (x, t ) = a sin (kx – ωt) + a sin (kx – ωt + φ )}

.........................(15.29)

color {green} { = a [ 2 sin [ ((kx - omega t ) + ( kx - omega t + phi ) )/2 ] cos \ \ phi/2]}

.................... (15.30)

color{blue} ✍️where we have used the familiar trignometric identity for sin A + sin B . We then have

color {blue } {y( x ,t) = 2a cos \ \phi/2 sin ( kx - omega t + phi/2 )}

....................(15.31)

color{blue} ✍️Eq. (15.31) 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 phi/2. The significant thing is that its amplitude is a function of the phase difference φ between the constituent two waves:

color{blue} {A(φ) = 2a cos ½φ}

.................. (15.32)

color{blue} ✍️For φ = 0, when the waves are in phase,

color{blue} {y (x,t ) = 2a sin (kx - omega t)}

.................. (15.33)

color{blue} ✍️i.e. the resultant wave has amplitude 2a, the largest possible value for A. For phi = pi ,

color{blue} ✍️the waves are completely, out of phase and the resultant wave has zero displacement everywhere at all times.

color{blue} {y (x, t ) = 0}

.................... (15.34)

color{blue} ✍️Eq. (15.33) refers to the so-called constructive interference of the two waves where the amplitudes add up in the resultant wave. Eq. (15.34) is the case of destructive intereference where the amplitudes subtract out in the resultant wave.

color{blue} ✍️Fig. 15.10 shows these two cases of interference of waves arising from the principle of superposition.

### REFLECTION OF WAVES

color{blue} ✍️So far we considered waves propagating in an unbounded medium. If a pulse or a wave meets a boundary and If the boundary is rigid, the pulse or wave gets reflected.

color{blue} ✍️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.

color{blue} ✍️A part of the incident wave is reflected and a part is transmitted into the second medium. If a wave is incident obliquely on the boundary between two different media the transmitted wave is called the "refracted wave."

color{blue} ✍️The incident and refracted waves obey Snell’s law of refraction, and the incident and reflected waves obey the usual laws of reflection.

color{blue} ✍️Fig. 15.11 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 180^0 on reflection.

color{blue} ✍️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.

color{blue} ✍️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.

color{blue} ✍️If on the other hend, the 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.

color{blue} ✍️The net maximum displacement at the boundary is then twice the amplitude of each pulse. An example of non- rigid boundary is the open end of an organ pipe.

color{blue} ✍️To summarize, a travelling wave or pulse suffers a phase change of π on reflection at a rigid boundary and no phase change on reflection at an open boundary. To put this mathematically, let the incident travelling wave be

color{ green} {y_2 (x,t ) = a sin (kx - omega t )}

color{blue} ✍️At a rigid boundary, the reflected wave is given by

color{blue} {y_r(x, t) = a sin (kx – ωt + π)}.
 color { blue} { = – a sin (kx – ωt)} ..............(15.35)

color{blue} ✍️At an open boundary, the reflected wave is given by

color {blue } {y_r(x, t) = a sin (kx – ωt + 0)}
color {blue } {= a sin (kx – ωt)} ................ (15.36)

color{blue} ✍️Clearly, at the rigid boundary, color{purple} { y = y_2 + y_r = 0 } at all times.