THE SPEED OF A TRAVELLING WAVE -

It should be noted that the speed of a mechanical wave is determined by the inertia and elastic properties of the medium. Inertia means mass density and elastic properties means Young’s modulus/ Bulk modulus/ Shear modulus.

When a part of the medium gets disturbed, it exerts an extra force on the neighbouring parts because of the elasticity property. The neighbouring parts respond to this force and the response depends on the inertia property.

To determine the speed of propagation of a travelling wave, we point out any particular point on the wave and see how that point moves in time.

Take a point P on a crest at time t, after a small time interval Δt the entire wave pattern is to shift to the right by a distance Δx. i.e point P on crest also move Δx in time Δt.

Consider the equation of travelling wave is $\dpi{120} \fn_cm y=A\sin(kx-\omega t)$

If the wave does not change its form while travelling. Then as time passes the phase of point P is constant.

i.e $\fn_cm kx-\omega t=constant$

Differentiating both sides w.r.t time, we get

$\dpi{120} \fn_cm \Rightarrow k \frac{dx}{dt}-\omega=0$

$\dpi{120} \fn_cm \Rightarrow \frac{dx}{dt}=\frac{\omega}{k}$

$\dpi{120} \fn_cm \left [ v=\frac{\omega}{k} \right ]$

Now, $\dpi{120} \fn_cm v=\frac{2\pi/T}{2\pi/\lambda}=\frac{\lambda}{T}$

$\dpi{120} \fn_cm \left [ v=f \lambda \right ]$

This is the general equation for all progressive waves. In a particular medium speed of the wave is constant. If we change the frequency of the wave then it automatically fixes the wavelength to maintain its speed.

SPEED OF TRANSVERSE WAVE ON STRETCHED STRING

The velocity of a wave travelling on a string depends on the elastic and the inertia properties of the string.

Consider a pulse travelling along a string with a speed v to the right. If the amplitude of the pulse is small compared to the length of the string, the tension T will be approximately constant along the string.

Let a reference frame moving with speed v to the right. looking from this frame, the pulse of the string is at rest but the entire string will pass through this crest in opposite direction like a snake with speed v.

Consider a small element of length Δl of the string at the highest point of a crest. Any small curve may be approximated by a circular arc of radius R(let). The particles of the string in this element go in this circle with a speed v. This element is pulled by the parts of the string to its right and to its left by tension.

from fig, $\dpi{120} \fn_cm T sin\theta+T \sin \theta=2 T sin \theta$ acts as a necessary centripetal force.

i.e $\dpi{120} \fn_cm 2T sin\theta=\frac{\Delta m \;v^2}{R}$

Where $\dpi{120} \fn_cm \Delta m\rightarrow$ mass of the small element

if $\dpi{120} \fn_cm \theta$ is small, then from fig

$\dpi{120} \fn_cm 2\theta=\frac{\Delta l}{R}\;\;\;\therefore \theta=\frac{\Delta l}{2R}$

$\dpi{120} \fn_cm Now, \;\;2T\theta=\frac{\Delta m\;v^2}{R}$

$\dpi{120} \fn_cm \Rightarrow 2 T.\frac{\Delta l}{2 R}=\frac{\Delta m\;v^2}{R}$

$\dpi{120} \fn_cm \therefore v^2=\frac{T}{\Delta m/ \Delta l}$

$\dpi{120} \fn_cm v^2=\frac{T}{\mu}\;\;\;\;\;or \;\;\;\;\;\left [ v=\sqrt{\frac{T}{\mu}} \right ]$

Where $\dpi{120} \fn_cm \mu\rightarrow$ mass per unit length

The velocity of wave on a string thus depends only on the tension T and linear mass density $\dpi{120} \fn_cm \mu$

SPEED OF TRANSVERSE WAVE

1. Speed of transverse wave in solid is given by

$\dpi{120} \fn_cm \left [ v=\sqrt{\frac{\eta}{\rho}} \right ]$

Where $\dpi{120} \fn_cm \eta\rightarrow$ modulus of rigidity of the material

and $\dpi{120} \fn_cm \rho\rightarrow$ density of the material

2. Speed of transverse waves in a stretched string is given by

$\dpi{120} \fn_cm \left [ v=\sqrt{\frac{T}{m}} \right ]$

Where $\dpi{120} \fn_cm T\rightarrow$ Tension

and $\dpi{120} \fn_cm m\rightarrow$ Mass per unit length of the string (linear density)

SPEED OF LONGITUDINAL WAVE

1. In solid

$\dpi{120} \fn_cm v=\sqrt{\frac{k+\frac{4}{3}\eta}{\rho}}$

Where $\dpi{120} \fn_cm k\rightarrow$ Bulk modulus, $\dpi{120} \fn_cm \eta\rightarrow$ Modulus of rigidity, $\dpi{120} \fn_cm \rho\rightarrow$ density of solid

2. In solid (In the form of rod i.e one dimension)

$\dpi{120} \fn_cm v=\sqrt{\frac{Y}{\rho}}$

Where $\dpi{120} \fn_cm Y\rightarrow$ Young’s modulus, $\dpi{120} \fn_cm \rho\rightarrow$ density of solid

3. In liquid

$\dpi{120} \fn_cm v=\sqrt{\frac{K}{\rho}}$

Where $\dpi{120} \fn_cm k\rightarrow$ Bulk modulus of liquid, $\dpi{120} \fn_cm \rho\rightarrow$ density of liquid

4. In gas

$\dpi{120} \fn_cm v=\sqrt{\frac{K}{\rho}}$

Where $\dpi{120} \fn_cm k\rightarrow$ Bulk modulus of gas, $\dpi{120} \fn_cm \rho\rightarrow$ density of gas

NOTE

1. Sound waves can travel through any material medium (solid, liquid or gases) with a speed that depends on the properties of the medium.

2. Sound travels faster in solids and liquids than in gas.

3. The compression and rarefaction causes only slight variations in air pressure (0.01 %)

4. The human ear or an electronic detector responds to the change in pressure and not to the displacement in a straightforward way.

5. Audible waves:- $\dpi{120} \fn_cm 20 Hz\leq f\leq 20kHz$

Inaudible waves:- Infrasonic waves $\dpi{120} \fn_cm < 20Hz$

Ultrasonic waves $\dpi{120} \fn_cm > 20KHz$

NEWTON’S FORMULA FOR VELOCITY OF SOUND

We know that the velocity of sound in air is given by

$\dpi{120} \fn_cm v=\sqrt{\frac{K}{\rho}}$

Where $\dpi{120} \fn_cm k\rightarrow$ Bulk modulus of air, $\dpi{120} \fn_cm \rho\rightarrow$ density of air

Newton assumed that when sound waves travel through the gas, the compressions and rarefactions are formed so slowly i.e the temperature of the medium remains the same (i.e isothermal process)

i.e $\dpi{120} \fn_cm v=\sqrt{\frac{k_{isothermal}}{\rho}}$

we know that

$\dpi{120} \fn_cm k=-\frac{dp}{(dv/v)}$

if the temperature is constant

$\dpi{120} \fn_cm pv=constant$

differentiate both sides, we get

$\dpi{120} \fn_cm pdv+vdp=0$

$\dpi{120} \fn_cm \therefore p=-\frac{dp}{(dv/v)}$

$\dpi{120} \fn_cm \left [ p=k_{isothermal} \right ]$

i.e velocity of sound is $\dpi{120} \fn_cm v=\sqrt{\frac{p}{\rho}}$

in NTP ( $\dpi{120} \fn_cm P=1\;atm=1.01\times10^5\;pa$ , $\dpi{120} \fn_cm \rho=1.293\; kg/m^3$ )

$\dpi{120} \fn_cm v=\sqrt{\frac{1.01\times10^5}{1.293}}=280 m/s$

But actual speed is 332 m/s (at NTP)

In 1817 Pierre Laplace explained the reason for this variation.

LAPLACE’S CORRECTION

According to Laplace, the compression and rarefaction are formed so rapidly that neither the heat is transferred to the surrounding. i.e process is adiabatic. (Gases are poor heat conductors)

i.e $\dpi{120} \fn_cm v=\sqrt{\frac{k_{adiabatic}}{\rho}}$

We know that

$\dpi{120} \fn_cm pv^{\gamma}=constant$

Differentiating both sides, we get

$\dpi{120} \fn_cm \Rightarrow \gamma\left ( pv^{\gamma-1} \right )dv+v^{\gamma}dp=0$

$\dpi{120} \fn_cm \Rightarrow dp=-\gamma \frac{pdv}{v}$

$\dpi{120} \fn_cm \therefore \gamma p=-\frac{dp}{(dv/v)}$

$\dpi{120} \fn_cm \left [ \gamma p=k_{adiabatic} \right ]$

i.e velocity of sound is $\dpi{120} \fn_cm v=\sqrt{\frac{\gamma p}{\rho}}$

For air $\dpi{120} \fn_cm \gamma =1.4$

at NTP

$\dpi{120} \fn_cm v=\sqrt{\frac{1.4\times 1.01 \times 10^5}{1.293}}$

$\dpi{120} \fn_cm \left [ v=331.2 m/s \right ]$

VELOCITY OF SOUND EFFECTING DUE TO VARIOUS FACTORS

1. Effect due to density:- If two gases having the same temperature and pressure and γ for both are the same.

If the densities of the gases be $\dpi{120} \fn_cm \rho_1\;and\; \rho_2$ , and speed of sound in this medium be $\dpi{120} \fn_cm v_1 \;and\; v_2$

Then $\dpi{120} \fn_cm v_1=\sqrt{\frac{\gamma p}{\rho_1}}\;\;\;\;\;and\;\;\;\;\;v_2=\sqrt{\frac{\gamma p}{\rho_2}}$

$\dpi{120} \fn_cm \therefore \frac{v_2}{v_1}=\sqrt{\frac{\rho_1}{\rho_2}}$

i.e the speed of sound in a gas is inversely proportional to the square root of the density.

2. Effect due to temperature:- For one mole of gas $\dpi{120} \fn_cm PV=RT$

If M is the molecular weight of the gas, then

$\dpi{120} \fn_cm \rho=\frac{M}{V}\;\;or\;\;V=\frac{M}{\rho}$

$\dpi{120} \fn_cm \therefore P\frac{M}{\rho}=RT\;\;\;\;\;or\;\;\;\;\;\frac{P}{\rho}=\frac{RT}{M}$

Now $\dpi{120} \fn_cm v=\sqrt{\frac{\gamma P}{\rho}}=\sqrt{\frac{\gamma RT}{M}}$

$\dpi{120} \fn_cm \therefore \left [ v\propto \sqrt{T} \right ]$

Hence, the velocity of sound in a gas is directly proportional to the square root of its absolute temperature.

3. Effect due to pressure:- We know that for one mole of gas $\dpi{120} \fn_cm PV=RT$

If T is constant, then

$\dpi{120} \fn_cm PV=constant$

$\dpi{120} \fn_cm \Rightarrow P\frac{M}{\rho}=constant$

$\dpi{120} \fn_cm \therefore \frac{P}{\rho}=constant$

We know that

$\dpi{120} \fn_cm v=\sqrt{\frac{\gamma P}{\rho}}=constant$

i.e the velocity of sound is independent of the pressure of the gas provided temperature remains constant.

4. Effect due to Humidity:- We know that $\dpi{120} \fn_cm v=\sqrt{\frac{\gamma p}{\rho}}$

The density of water vapour is less than dry air at the same pressure. Thus, the density of moist air is less than that of dry air. As a result, the speed of sound increases with increasing humidity.