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Chapter: 11th Physics : Waves

Velocity of Waves in Different Media

Suppose a hammer is stroked on long rails at a distance and when a person keeps his ear near the rails at the other end he/she will hear two sounds, at different instants.

VELOCITY OF WAVES IN DIFFERENT MEDIA

Suppose a hammer is stroked on long rails at a distance and when a person keeps his ear near the rails at the other end he/she will hear two sounds, at different instants. The sound that is heard through the rails (solid medium) is faster than the sound we hear through the air (gaseous medium). This implies the velocity of sound is different in different media.

In this section, we shall derive the velocity of waves in two different cases:

1. The velocity of a transverse waves along a stretched string.

2. The velocity of a longitudinal waves in an elastic medium.

 

1. Velocity of transverse waves in a stretched string

 

Let us compute the velocity of transverse travelling waves on a string. When a jerk is given at one end (left end) of the rope, the wave pulses move towards right end with a velocity v as shown in the Figure 11.15 (a). This means that the pulses move with a velocity v with respect to an observer who is at rest frame. Suppose an observer also moves with same velocity v in the direction of motion of the wave pulse, then that observer will notice that the wave pulse is stationary and the rope is moving with pulse with the same velocity v.


Consider an elemental segment in the string as shown in the Figure 11.15 (b). Let A and B be two points on the string at an instant of time. Let dl and dm be the length and mass of the elemental string, respectively. By definition, linear mass density, μ is


The elemental string AB has a curvature which looks like an arc of a circle with centre at O, radius R and the arc subtending an angle θ at the origin O as shown in Figure 11.15(b). The angle θ can be written in terms of arc length and radius as θ = dl/R. The centripetal acceleration supplied by the tension in the string is


Then, centripetal force can be obtained when mass of the string (dm) is included in equation (11.7)


The centripetal force experienced by elemental string can be calculated by substituting equation (11.6) in equation (11.8) we get


The tension T acts along the tangent of the elemental segment of the string at A and B. Since the arc length is very small, variation in the tension force can be ignored. We can resolve T into horizontal component Tcos(θ/2) and vertical component T sin(θ/2).

The horizontal components at A and B are equal in magnitude but opposite in direction; therefore, they cancel each other. Since the elemental arc length AB is taken to be very small, the vertical components at A and B appears to acts vertical towards the centre of the arc and hence, they add up. The net radial force Fr is


Since the amplitude of the wave is very small when it is compared with the length of the string, the sine of small angle is approximated as sin(θ/2)≈ θ/2. Hence, equation (11.10) can be written as

But θ =  dl/R, therefore substituting in equation (11.11), we get


Applying Newton’s second law to the       elemental string in the radial direction, under equilibrium, the radial component of the force is equal to the centripetal force. Hence equating equation (11.9) and equation (11.12), we have


 

Observations:

The velocity of the string is

a. directly proportional to the square root of the tension force

b. inversely proportional to the square root of linear mass density

c. independent of shape of the waves.

 

EXAMPLE 11.6

Calculate the velocity of the travelling pulse as shown in the figure below. The linear mass density of pulse is 0.25 kg m-1. Further, compute the time taken by the travelling pulse to cover a distance of 30 cm on the string.


Solution

The tension in the string is T = m g = 1.2 × 9.8 = 11.76 N

The mass per unit length is μ = 0.25 kg m-1. Therefore, velocity of the wave pulse is


The time taken by the pulse to cover the distance of 30 cm is


where

ms = milli second.

 

2. Velocity of longitudinal waves in an elastic medium

 

Consider an elastic medium (here we assume air) having a fixed mass contained in a long tube (cylinder) whose cross sectional area is A and maintained under a pressure P. One can generate longitudinal waves in the fluid either by displacing the fluid using a piston or by keeping a vibrating tuning fork at one end of the tube. Let us assume that the direction of propagation of waves coincides with the axis of the cylinder. 


Let ρ be the density of the fluid which is initially at rest. At t = 0, the piston at left end of the tube is set in motion toward the right with a speed u.

Let u be the velocity of the piston and v be the velocity of the elastic wave. In time interval  ∆t, the distance moved by the piston ∆d = ut. Now, the distance moved by the elastic disturbance is  ∆x = vt. Let  m be the mass of the air that has attained a velocity v in a time         ∆t . Therefore,

m = ρ Ax = ρ A (vt)

Then, the momentum imparted due to motion of piston with velocity u is

p = [ρ A (v  t)]u

But the change in momentum is impulse.

The net impulse is


When the sound wave passes through air, the small volume element (ΔV) of the air undergoes regular compressions and rarefactions. So, the change in pressure can also be written as


where, V is original volume and B is known as bulk modulus of the elastic medium.

But V = Ax = A vt and

V = Ad =A ut

Therefore,


Comparing equation (11.14) and equation (11.15), we get


In general, the velocity of a longitudinal wave in elastic medium is v = √E/√P where E is the modulus of elasticity of the medium.

 

Cases: For a solid :

(i) one dimension rod (1D)


where Y is the Young’s modulus of the material of the rod and ρ is the density of the rod. The 1D rod will have only Young’s modulus.

(ii)Three dimension rod (3D) The speed of longitudinal wave in a solid is


where η is the modulus of rigidity, K is the bulk modulus and ρ is the density of the rod.

Cases: For liquids:


where, K is the bulk modulus and ρ is the density of the rod.

 

EXAMPLE 11.7

Calculate the speed of sound in a steel rod whose Young’s modulus Y = 2 × 1011 N m-2 and ρ = 7800 kg m-3.

Solution


Therefore, longitudinal waves travel faster in a solid than in a liquid or a gas. Now you may understand why a shepherd checks before crossing railway track by keeping his ears on the rails to safegaurd his cattle.

 

EXAMPLE 11.8

An increase in pressure of 100 kPa causes a certain volume of water to decrease by 0.005% of its original volume.

(a) Calculate the bulk modulus of water?.

(b) Compute the speed of sound (compressional waves) in water?.

Solution

(a) Bulk modulus


(b) Speed of sound in water is




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