A bullet passes past a person at a speed $220\,m{s^{ - 1}}$. The fractional change in the frequency of the whistling sound heard by the person as the bullet crosses the person is (Speed of sound $ = 330\,m{s^{ - 1}}$ ):
A. $0.67$
B. $0.8$
C. $1.2$
D. $3.0$
Answer
635.4k+ views
Hint:Here we have to use the principle of Doppler Effect of sound. The Doppler effect is the difference in a wave’s frequency in response to an observer that shifts relative to the source of the wave. The change of pitch heard as a vehicle with a horn approaches and retreats from an observer is a typical example of Doppler Effect.
Complete answer:
Let the original frequency of the source be ${f_ \circ }$.
Speed of source, ${V_{source}} = 220\,m{s^{ - 1}}$.
When the bullet moves towards the observer the Doppler effect is:
Let the apparent frequency be $f'$.
$
f' = {f_ \circ }\left[ {\dfrac{{{V_{sound}}}}{{{V_{sound}} - {V_{source}}}}} \right] \\
\Rightarrow f' = {f_ \circ }\left[ {\dfrac{{330}}{{330 - 220}}} \right] \\
\Rightarrow f' = 3{f_ \circ } \\$
When the bullet moves away the observer the Doppler effect is:
Let the apparent frequency be ${f^{11}}$.
$
{f^{11}} = {f_ \circ }\left[ {\dfrac{{330}}{{330 - 220}}} \right] \\
\Rightarrow{f^{11}} = 3{f_ \circ } \\ $
Hence, the fractional change in frequency is: $\dfrac{{f' - {f^{''}}}}{{f'}}$
$\dfrac{{f' - {f^{''}}}}{{f'}}\\
\Rightarrow\dfrac{{3{f_ \circ } - 0.6{f_ \circ }}}{{3{f_ \circ }}} \\
\therefore 0.8 \\ $
Hence, option B is correct.
Additional information:
The Doppler Effect is caused when waves are transmitted out at a normal rate or frequency by the source of a waveform, such as sound or light, but there is continuous relative motion between the source and the observer, allowing frequency measured to change. The explanation for Doppler Effect is that each subsequent wave crest is produced from a position closer to the observer than the crest of the previous wave as the wave source travels towards the observer. Thus, each wave takes significantly less time than the previous wave to enter the observer.
Note:Here we have to see what the speed is when the bullet passes the person. Then only we can apply the Doppler effect. Also we have to correctly remember the formula of the Doppler Effect otherwise the answer will be wrong.
Complete answer:
Let the original frequency of the source be ${f_ \circ }$.
Speed of source, ${V_{source}} = 220\,m{s^{ - 1}}$.
When the bullet moves towards the observer the Doppler effect is:
Let the apparent frequency be $f'$.
$
f' = {f_ \circ }\left[ {\dfrac{{{V_{sound}}}}{{{V_{sound}} - {V_{source}}}}} \right] \\
\Rightarrow f' = {f_ \circ }\left[ {\dfrac{{330}}{{330 - 220}}} \right] \\
\Rightarrow f' = 3{f_ \circ } \\$
When the bullet moves away the observer the Doppler effect is:
Let the apparent frequency be ${f^{11}}$.
$
{f^{11}} = {f_ \circ }\left[ {\dfrac{{330}}{{330 - 220}}} \right] \\
\Rightarrow{f^{11}} = 3{f_ \circ } \\ $
Hence, the fractional change in frequency is: $\dfrac{{f' - {f^{''}}}}{{f'}}$
$\dfrac{{f' - {f^{''}}}}{{f'}}\\
\Rightarrow\dfrac{{3{f_ \circ } - 0.6{f_ \circ }}}{{3{f_ \circ }}} \\
\therefore 0.8 \\ $
Hence, option B is correct.
Additional information:
The Doppler Effect is caused when waves are transmitted out at a normal rate or frequency by the source of a waveform, such as sound or light, but there is continuous relative motion between the source and the observer, allowing frequency measured to change. The explanation for Doppler Effect is that each subsequent wave crest is produced from a position closer to the observer than the crest of the previous wave as the wave source travels towards the observer. Thus, each wave takes significantly less time than the previous wave to enter the observer.
Note:Here we have to see what the speed is when the bullet passes the person. Then only we can apply the Doppler effect. Also we have to correctly remember the formula of the Doppler Effect otherwise the answer will be wrong.
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