Question

The frequency of a whistle of an engine is 600 cycles / sec is moving with the speed of 30 m / sec towards an observer. The apparent frequency will be (velocity of sound = 330 m / s )

Answer 1

Answer:

Given:

Actual frequency of whistle = 600 Hz

Speed of source = 30 m/s

Velocity of observer = 0 m/s

To find:

Apparent frequency heard by the observer.

Concept:

When an source moves towards the observer, the observer actually hears a sound with an apparently higher Frequency as compared to the source.

This effect is known as Doppler's Effect.

This effect is also seen in Electromagnetic waves like that of light.

Calculation:

According to Doppler's Effect :

$f2 = \bigg \{ \dfrac{v - v_{o}}{v - v_{s}} \bigg \}f$

$= > f2 = \bigg \{ \dfrac{330 - 0}{330 - 30} \bigg \} \times 600$

$= > f2 = \bigg \{ \dfrac{330 }{300} \bigg \} \times 600$

$= > f2 = 660\: hz$

So final answer :

$\boxed{ \huge{ \sf{ \green{ \bold{f2 = 660\: hz}}}}}$

Answer 2

$\huge \underline {\underline{ \mathfrak{ \green{Ans}wer \colon}}}$

Given :

• Frequency of whistle (f) = 600 Hz
• Speed of Observer (Vo) = 0 m/s
• Speed of Source (Vs) = 30 m/s

________________________

To Find :

• Apparent Frequency of Whistle

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Solution :

As we know that from Doppler's Effect that :

$\large{\boxed{\sf{f' \: = \: \bigg( \dfrac{V \: - \: V_o}{V \: - \: V_s} \bigg) f}}} \\ \\ \implies {\sf{f' \: = \: \bigg( \dfrac{330 \: - \: 0}{330 \: - \: 30}\bigg) \: \times \: 600}} \\ \\ \implies {\sf{f' \: = \: \dfrac{330}{300} \: \times \: 600}} \\ \\ \implies {\sf{f' \: = \: 330 \: \times \: 2}} \\ \\ \implies {\sf{f' \: = \: 660}} \\ \\ \underline{\sf{\therefore \: Apparent \: frequency \: is \: 660 \: Hz}}$