Definition & Proof Based Problems

Q 3176891776

Define kinetic energy density of a wave. Derive an expression for its maximum value. Using it, prove that the intensity of the wave is, `I = 1/2 pomega^2A^2`

For a travelling wave, `y = A sin(omegat - Kx),` be the displacement. The velocity of particles is

given by `v+p = (dy/dt) = omegaA cos(omegat -Kx)`

K.E. per unit volume

`1/2 m/v v^2 = 1/2 pv^2`

:. K.E. density defined as K.E. per unit

volume is given by `1/2 pomega^2 A^2 cos^2 (omegat-Kx)`

Maximum value of energy density

`= 1/2 pomega^2A^2`

Intensity is the energy falling per unit area per unit time.

So, `I = (DeltaE)/(DeltatS) = P/S`

`I = (1/2pomega^2 A^2 . SDeltax)/(SDeltat) =1/2 pomega^2A^2v`

Q 3176191076

Define temperature coefficient of the velocity of sound.

The temperature coefficient of the velocity of sound is defined as the change in the velocity of sound, when the temperature changes by `1°C` (or 1 K). It is denoted by `alpha`.

Q 3106891778

A wave `Y =A sin{omegat- kx)` allowed through a string gets reflected from a rigid support and forms stationary wave. Derive the expression for the standing wave.

Given wave : `Y_i =A sin(omegat- Kx)`

Reflected wave :

`Y_r = A sin(omegat +Kx +pi)`

`= - A sin(omegat+Kx)`

Q 3116891779

List the differences between a progressive and a stationary wave.

Q 3167101085

What are standing waves ? Discuss graphical method for formation of standing waves on stretched strings.

Standing waves are the pattern of waves produced when two waves moving in opposite direction interact. They do not transport energy.

If a wave 'A' is made to hit a rigid support, there will be a reflected wave 'B' from the rigid support. As the two waves are super imposed, there will be a standing wave (as shown) produced. It can be represented by the equation, `y = 2A sin kx` cos wt indicating a position varying amplitude `(2A sin kx)`

Q 3147401383

A source of sound is coming towards a stationary observer with a constant velocity. Deduce the formula for the apparent frequency heard by observer.

Let S and 0 be the source and observer. If v is the frequency of sound with velocity v released by the source, then number of waves will be received by the observer at rest. When the source approaches the stationary observer, the number of waves received increases due to the apparent shortening of the wavelength. Wavelength perceived

`lambda= (velocity of sound w.r.t. moving source)/(frequency)`

`: . lambda = (u-u_s)/v`

using `lambda= l/v` we get

and `u/v = (v-v_s)/v ; v' = v (v/(v-v_s))`

Q 3136691572

Obtain an expression for apparent frequency of sound. when the source is moving with a velocitJ `v_5` towards the stationary listener.

Let u be the velocity of sound and `u_8` be the velocity of source towards observer. As the source approaches, the space between source and observer gets reduced but should accomodate the same number of waves (as frequency). So, wavelength reduces. The new waveiength is,

`lambda' = text(velocity of sound w.r.t. source)/text(frequency)`

`= (v-v_s)/v`

Also `lambda = u/v`

`v' =v (v/v-v_s)`

Q 3137301282

Explain Doppler effect in sound. Obtain an expression for apparent frequency of sound when source and listener are approaching each other.

Whenever there is a relative motion between a source of sound and listener, the apparent frequency of sound heard by listener is different from actual frequency of sound emitted by the source. Let S be a source of sound and L, the listener of sound, both initially at rest. Let u be the actual frequency of sound emitted by the source and A be the actual wavelength of sound emitted. If u is velocity of sound in still air, then `lambda= u/v`

Let the distance between source and listener be V, so that v waves from the source reach listener in 1 second. `V, =` Velocity of medium,

`V_s =` Velocity of source, `V_L =` Velocity of listener

Resultant velocity of sound along

`SL = (V + V_m)`

SS' = distance proved by source in 1 sec.

`= V_s along SL`

. . Relative velecity of sound w.r.t. source

`= f(V +V_m)- V_sl`

As the frequency remains unchanged.

V waves emitted in one second occupy

the distance `[(V + V_m) - V_sl`

`lambda = [(V+V_m -V_s]`

`LL' = VL` Relative vel. of sound waves w.r.t listener `(V + Vm) - VL`

Apparent frequency of sound waves heard by listener is

`v' ((V+B_m)- V_L)/lambda`

`v' ( [(V+B_m)- V_L)/(V+ V_m )-V_S]) =V`

when both approach each other

`V_S = (+), V_L = (-)`

`V' = (V- (-V_L))/(V-V_S) V = (V+V_L)/(V-V_S) )V`

Q 3186491377

Obtain an expression for apparent frequency of sound. when the source is moving with a velocitJ `v_s` towards the stationary listener.

Let u be the velocity of sound and `u_8` be the velocity of source towards observer. As the source approaches, the space between source and observer gets reduced but should accomodate the same number of waves (as frequency). So, wavelength reduces. The new waveiength is,

`lambda' = text(velocity of sound w.r.t. source)/text(frequency)`

`= (v-v_s)/v`

Also `lambda = u/v`

`v' =v (v/v-v_s)`