Question

Question 8 Consider a source emitting sound at frequency f. The speed of sound wave in the medium is v. (a) Show that the Doppler frequency is for a stationary observer as the source approaches at speed v 12 marks (b) Show that the Doppler frequency is f for a stationary observer as the source recedes at speed v, [2 marks (c) Show that the Doppler frequency is vv as the observer and the source are travelling towards each other at speeds v, and v, respectively, relative to the fixed ground. [2 marks A stationary pedestrian is hearing a frequency of 550 Hz from the siren of an approach- ing police car and a frequency of 450 Hz as the police car has passed and is receding. The speed of sound in air is v 343 ms (d) How fast are the police travelling? [2 marks] (e) The police siren is radiating 100 W of sound energy in all directions. As the police approach the pedestrian, at a distance of 50 m: [2 marks () What is the sound intensity level that the pedestrian is hearing? (i) What is the rate at which the sound intensity level is changing? A driver is driving at a speced of 50 km h in the opposite direction of the police car. As the police travel at the speed cal culated in Question 8(d) towards the [3 marks driver, at a distance of 50 m: (i) What is the sound frequency that the driver is hearing? (i) What is the sound intensity level that the driver is hearing? (ii) What is the rate at which the sound intensity level is changing?

Answer 1

(a) Velocity of sound of source relative to stationary observer = $\nu$ - $\nu$_s ..........(1) when source is approaching therefore,

$\lambda$ =( $\nu$ - $\nu$_s) / f ..............(2)

if f₊ is the frequency of sound heard by the listener then $\nu$ = f₊$\lambda$ ........(2)

on equating (1) and (2) we get

$\nu$/f₊ = ($\nu$ - $\nu$_s )/ f ..........(3)

or, f₊ = $\nu$f/($\nu$ - $\nu$_s)

(b) Velocity of sound of source relative to stationary observer = $\nu$ + $\nu$_s ..........(1) therefore,

$\lambda$ =( $\nu$ + $\nu$_s) / f ..............(2)

if f₊ is the frequency of sound heard by the stationary observer then $\nu$ = f₊$\lambda$ ........(2)

on equating (1) and (2) we get

$\nu$/f₊ = ($\nu$ + $\nu$_s )/ f ..........(3)

or, f₊ = $\nu$f/($\nu$ + $\nu$_s

(c) when both are coming closer ,

wavelength of sound heard by the observer = ( $\nu$ + $\nu$_o )/ f₊ .........(3)

wavelength of sound heard by the source =($\nu$ - $\nu$_s)/f .............(4)

thus from (3) and (4) we get (v - v_s )f₊ = (v + v_o)f

or f₊ =( v + v_o)f/ (v - v_s)

(d) 34.3Hz

solution:

let speed of the car = c m/s

then

when police car is approaching frequency heard by stationary pedestrian is

550 = nv/v - v_{s   ...(1)}

when police car is receding frequency heard by stationary pedestrian is

450 = nv/v + v_{s ........(2)}

_{from (1) and (2) we get}

550(v - v_s) = 450(v + v_s)

or , v_s = 100v/1000 = 343/10 = 34.3Hz