A source of sound is moving with constant velocity of $20\, m/s$ emitting a note of frequency $1000 \,Hz.$ The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be
(Speed of sound $v = 340\, m/s$)
Medium
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(a) When source is approaching the observer, the frequency heard
${n_a} = \left( {\frac{v}{{v - {v_S}}}} \right) \times n = \left( {\frac{{340}}{{340 - 20}}} \right) \times 1000 = 1063Hz$
When source is receding, the frequency heard
${n_r} = \left( {\frac{v}{{v + {v_S}}}} \right) \times n$=$\frac{{340}}{{340 + 20}} \times 1000 = 944$
$ \Rightarrow {n_a}:{n_r} = 9:8$
Short trick : $\frac{{{n_a}}}{{{n_r}}} = \frac{{v + {v_S}}}{{v - {v_S}}} = \frac{{340 + 20}}{{340 - 20}} = \frac{9}{8}.$
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