Question

1. A source producing a sound of frequency 90Hz is approaching a stationary listener with a speed equal to (1/10) of the speed of sound. What will be the frequency heard by the listener ?
2. A source producing a sound of frequency 500 Hz is moving towards a listener with a velocity of 30 ms^-1. what will be the frequency heard by listener ?
3. A source producing a sound of frequency 500 Hz is moving towards a listener with a velocity of 30 ms^-1. what will be the frequency heard by listener ?

Answer 1

Answer:

100 Hz

Explanation:

1. n' =V/V-Vs n

We are given n =90 Hz and Vs (1/10)

Substituting these values in the equation of Doppler's effect in sound.

n' = V/V-(1/10)V 90= 100Hz

2. Given; n=500Hz,V

s

=330m/s,v=30m/s,n

′

=?

n′=n( v−v sv)

⇒n′=500(300−30

330)=550 Hz

3. same as 2nd one.

Please mark me as the brainliest.

Answer 2

1. When the source is moving towards the stationary listener, the expression for apparent frequency is

$n' = ( \frac{V}{V-V_s}) n$

$= (\frac{V}{V - (\frac{1}{10}) v })n$

$= (\frac{V}{V- \frac{V}{10}}) n$

$= (\frac{V}{\frac{10V-V}{10}}) n$

$= (\frac{V}{\frac{9V}{10}}) n \implies( \frac{10V}{9V})n$

$= (\frac{10}{9})n} \implies (\frac{10}{9}) \times 90$

$\boxed{100\:\text{Hz}}$

2. When the source is moving towards the stationary listener, the expression for apparent frequency is

$n' = (\frac{V}{V - V_s})n$

$n' = (\frac{330}{ 330 -30}) n$

$= (\frac{330}{300}) \times 500$

$= (\frac{11}{10}) \times 500$

$\boxed{550 \: \text{Hz}}$

3. When the source is moving towards the stationary listener, the expression for apparent frequency is

$n' = (\frac{V}{V - V_s})n$

$n' = (\frac{330}{ 330 -30}) n$

$= (\frac{330}{300}) \times 500$

$= (\frac{11}{10}) \times 500$

$= (\frac{11}{10}) \times 500$

$\boxed{550 \: \text{Hz}}$