A source of sound is moving with a constant v | vedclass

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(A) When the source is approaching the stationary observer,the observed frequency $n_a$ is given by the Doppler effect formula:$n_a = \left( \frac{v}{

Vedclass · Accepted Answer

$9:8$

Vedclass · Answer

(A) When the source is approaching the stationary observer,the observed frequency $n_a$ is given by the Doppler effect formula:$n_a = \left( \frac{v}{v - v_s} ight) n$Substituting the values $v = 340\, m/s$,$v_s = 20\, m/s$,and $n = 1000\, Hz$:$n_a = \left( \frac{340}{340 - 20} ight) 	imes 1000 = \frac{340}{320} 	imes 1000 = \frac{17}{16} 	imes 1000 = 1062.5\, Hz$When the source is receding from the observer,the observed frequency $n_r$ is:$n_r = \left( \frac{v}{v + v_s} ight) n$$n_r = \left( \frac{340}{340 + 20} ight) 	imes 1000 = \frac{340}{360} 	imes 1000 = \frac{17}{18} 	imes 1000 = 944.4\, Hz$The ratio of the frequencies is:$\frac{n_a}{n_r} = \frac{\frac{v}{v - v_s} n}{\frac{v}{v + v_s} n} = \frac{v + v_s}{v - v_s}$$\frac{n_a}{n_r} = \frac{340 + 20}{340 - 20} = \frac{360}{320} = \frac{9}{8}$Thus,the ratio is $9:8$.

$A$ source of sound is moving with a 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 $v = 340\, m/s$).

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