
A source of sound of frequency 450 cycles/sec is moving towards a stationary observer with 34m/s speed. If the speed of sound is 340 m/s, then the apparent frequency will be
A. 410 cycles/sec
B. 500 cycles/sec
C. 550 cycles/sec
D. 450 cycles/sec
Answer
580.2k+ views
Hint: When there is relative motion between the source of sound and the observer, the frequency of the sound heard by the observer is different from the actual frequency of the sound. Use the formula for the apparent frequency when the source of sound is moving towards a stationary observer to find the apparent frequency.
Formula used: ${{f}^{'}}=f\left( \dfrac{v}{v-{{v}_{s}}} \right)$
Complete step by step answer:
It is given that a source of sound is moving towards a stationary observer. The frequency of the sound emitted by the source is 450 cycles/sec and the source is miving with a speed of 34m/s. It is also given that the speed of sound in this medium is 340 m/s.
When there is relative motion between the source of sound and the observer, the frequency of the sound heard by the observer is different from the actual frequency of the sound (i.e. the frequency of the sound that is heard when the observer and the source are at rest).
When the source moves towards a stationary observer, the apparent frequency of the sound is given as ${{f}^{'}}=f\left( \dfrac{v}{v-{{v}_{s}}} \right)$ …… (i),
where f is the actual frequency, v is the speed of the sound and ${{v}_{s}}$ is the speed of the source.
In this case, f = 450 cycles/sec, v = 340m/s and ${{v}_{s}}=34m{{s}^{-1}}$.
Substitute the values in (i).
${{f}^{'}}=450\left( \dfrac{340}{340-34} \right)=450\times \dfrac{340}{306}=500cycles/\sec $.
So, the correct answer is “Option B”.
Note: From this solution, we can understand that when the source of sound moves towards a stationary observer, the frequency of the sound heard by the observer is more than the actual frequency of the sound.
On the other hand, when the source moves away from the stationary observer, the apparent frequency is less than the actual frequency.
Formula used: ${{f}^{'}}=f\left( \dfrac{v}{v-{{v}_{s}}} \right)$
Complete step by step answer:
It is given that a source of sound is moving towards a stationary observer. The frequency of the sound emitted by the source is 450 cycles/sec and the source is miving with a speed of 34m/s. It is also given that the speed of sound in this medium is 340 m/s.
When there is relative motion between the source of sound and the observer, the frequency of the sound heard by the observer is different from the actual frequency of the sound (i.e. the frequency of the sound that is heard when the observer and the source are at rest).
When the source moves towards a stationary observer, the apparent frequency of the sound is given as ${{f}^{'}}=f\left( \dfrac{v}{v-{{v}_{s}}} \right)$ …… (i),
where f is the actual frequency, v is the speed of the sound and ${{v}_{s}}$ is the speed of the source.
In this case, f = 450 cycles/sec, v = 340m/s and ${{v}_{s}}=34m{{s}^{-1}}$.
Substitute the values in (i).
${{f}^{'}}=450\left( \dfrac{340}{340-34} \right)=450\times \dfrac{340}{306}=500cycles/\sec $.
So, the correct answer is “Option B”.
Note: From this solution, we can understand that when the source of sound moves towards a stationary observer, the frequency of the sound heard by the observer is more than the actual frequency of the sound.
On the other hand, when the source moves away from the stationary observer, the apparent frequency is less than the actual frequency.
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