
A source producing the sound of frequency 170 Hz is approaching a stationary observer with a velocity of 17 m/s. The apparent change in the wavelength of sound heard by the observer is (speed of sound in air = 340 m/s)
A. 0.1m
B. 0.2m
C. 0.4m
D. 0.5m
Answer
164.4k+ views
Hint:To find the apparent change in wavelength, we find the apparent change in the frequency due to Doppler effect and then using the relation between the wave speed, frequency and the wavelength, we get the apparent change in wavelength of the sound.
Formula used:
\[{f_{ap}} = {f_o}\left( {\dfrac{{v \pm {v_o}}}{{v \pm {v_s}}}} \right)\]
where \[{f_{ap}}\] is the apparent frequency heard by the listener moving with speed \[{v_o}\] with respect to the source which is moving with speed \[{v_s}\], \[{f_o}\] is the original frequency and v is the speed of sound in air.
\[v = f\lambda \]
where v is the speed of the wave, f is the frequency of the wave and \[\lambda \] is the wavelength of the wave.
Complete step by step solution:
It is given that the source of the sound is approaching towards the stationary observer.
The Doppler Effect formula for the sound source moving towards the stationary observer is,
\[{f_{ap}} = {f_o}\left( {\dfrac{v}{{v - {v_s}}}} \right)\]
The speed of sound, v is given as 340 m/s and the original frequency is given as 170 Hz.
So, the apparent frequency is,
\[{f_{ap}} = \left( {170\,Hz} \right)\left( {\dfrac{{340}}{{340 - 17}}} \right) \\ \]
\[\Rightarrow {f_{ap}} = 170 \times \dfrac{{340}}{{323}}Hz \\ \]
\[\Rightarrow {f_{ap}} = 179\,Hz\]
Using the relation between the speed of the wave, frequency and the wavelength we get the initial and final wavelength. By finding differences, we get the change in wavelength.
\[\Delta \lambda = {\lambda _i} - {\lambda _f} \\ \]
\[\Rightarrow \Delta \lambda = \dfrac{v}{{{f_o}}} - \dfrac{v}{{{f_{ap}}}} \\ \]
\[\Rightarrow \Delta \lambda = \left( {\dfrac{{340}}{{170}} - \dfrac{{340}}{{179}}} \right)m \\ \]
\[\therefore \Delta \lambda = 0.10\,m\]
Hence, the apparent change in the wavelength of the sound wave is 0.10 m.
Therefore, the correct option is A.
Note: The Doppler’s effect is for the change in the frequency due to relative motion between the source of the sound and the observer. So, we can’t directly use the Doppler’s effect for the change in wavelength for the mechanical wave-like sound.
Formula used:
\[{f_{ap}} = {f_o}\left( {\dfrac{{v \pm {v_o}}}{{v \pm {v_s}}}} \right)\]
where \[{f_{ap}}\] is the apparent frequency heard by the listener moving with speed \[{v_o}\] with respect to the source which is moving with speed \[{v_s}\], \[{f_o}\] is the original frequency and v is the speed of sound in air.
\[v = f\lambda \]
where v is the speed of the wave, f is the frequency of the wave and \[\lambda \] is the wavelength of the wave.
Complete step by step solution:
It is given that the source of the sound is approaching towards the stationary observer.
The Doppler Effect formula for the sound source moving towards the stationary observer is,
\[{f_{ap}} = {f_o}\left( {\dfrac{v}{{v - {v_s}}}} \right)\]
The speed of sound, v is given as 340 m/s and the original frequency is given as 170 Hz.
So, the apparent frequency is,
\[{f_{ap}} = \left( {170\,Hz} \right)\left( {\dfrac{{340}}{{340 - 17}}} \right) \\ \]
\[\Rightarrow {f_{ap}} = 170 \times \dfrac{{340}}{{323}}Hz \\ \]
\[\Rightarrow {f_{ap}} = 179\,Hz\]
Using the relation between the speed of the wave, frequency and the wavelength we get the initial and final wavelength. By finding differences, we get the change in wavelength.
\[\Delta \lambda = {\lambda _i} - {\lambda _f} \\ \]
\[\Rightarrow \Delta \lambda = \dfrac{v}{{{f_o}}} - \dfrac{v}{{{f_{ap}}}} \\ \]
\[\Rightarrow \Delta \lambda = \left( {\dfrac{{340}}{{170}} - \dfrac{{340}}{{179}}} \right)m \\ \]
\[\therefore \Delta \lambda = 0.10\,m\]
Hence, the apparent change in the wavelength of the sound wave is 0.10 m.
Therefore, the correct option is A.
Note: The Doppler’s effect is for the change in the frequency due to relative motion between the source of the sound and the observer. So, we can’t directly use the Doppler’s effect for the change in wavelength for the mechanical wave-like sound.
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