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# Can someone explain A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 ms-1 at an angle of 60° with the source observer line as shown in

A source of sound S emitting waves of frequency 100 Hz and an observer O are located at some distance from each other. The source is moving with a speed of 19.4 ms-1 at an angle of 60° with the source observer line as shown in the figure. The observer is at rest. The apparent frequency observed by the observer (velocity of sound in air 330 ms-1), is:

• Option 1)

103 Hz

• Option 2)

106 Hz

• Option 3)

97 Hz

• Option 4)

100 Hz

Answers (1)
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As discussed in

frequency of sound when source is moving towards the observer and observer is moving away from source -

$\nu {}'= \nu _{0}.\frac{C-V_{0}}{C-V_{s}}$

- wherein

$C=$ Speed of sound

$V_{0}=$ Speed of observer

$V_{s}=$ speed of source

$\nu _{0 }=$ Original frequency

$\nu {}'=$ apparent frequency

From Doppler's effect,

$\nu '= \nu _{0}\left ( \frac{v-v_{0}}{v-v_{s}} \right )$

$\nu '= 100 \left (\frac{v-0}{v - \left ( +9.7 \right )} \right )$

$\Rightarrow \nu '= \frac{100v}{v\left ( 1- \frac{9.7}{v} \right )}$

$= \frac{100}{\left ( 1-\frac{9.7}{330} \right )}$

$\nu' = 103 Hz$

Option 1)

103 Hz

This option is correct

Option 2)

106 Hz

This option is incorrect

Option 3)

97 Hz

This option is incorrect

Option 4)

100 Hz

This option is incorrect

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