wave_mechanics dopplers_effect Question 64
Question: The apparent frequency of a note, when a listener moves towards a stationary source, with velocity of 40 m/s is 200 Hz. When he moves away from the same source with the same speed, the apparent frequency of the same note is 160 Hz. The velocity of sound in air is (in m/s)
[KCET 1998]
Options:
360
330
320
340
Show Answer
Answer:
Correct Answer: A
Solution:
When a listener moves towards a stationary source apparent frequency $ {n}’=( \frac{v+v _{0}}{v} ) $
$ n=200 $ ?..(i) When listener moves away from the same source $ {n}’’=\frac{(v-v _{L})}{v}n=160 $ ?..(ii) From (i) and (ii) $ \frac{v+v _{L}}{v-v _{L}}=\frac{200}{160} $
Therefore $ \frac{v+v _{O}}{v-v _{O}}=\frac{5}{4} $
Therefore $ v=360\ \text{m}/\sec $
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