A source of sound is moving with constant velocity of $20\text{ ms}^{-1}$ emitting a note of frequency…
A source of sound is moving with constant velocity of $20\text{ ms}^{-1}$ emitting a note of frequency $1000\text{ Hz}$. 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, $v = 340\text{ ms}^{-1}$)
$9 : 8$
$8 : 9$
$1 : 1$
$9 : 10$
Solution
When source is approaching the observer, the frequency heard,
$f_a = \left( \frac{v}{v - v_s} \right) \times f = \left( \frac{340}{340 - 20} \right) \times 1000$
$= 1062.5\text{ Hz} \approx 1062\text{ Hz}$
When source is receding, the frequency heard,
$f_r = \left( \frac{v}{v + v_s} \right) \times f = \left( \frac{340}{340 + 20} \right) \times 1000 = 944.4\text{ Hz} \approx 944\text{ Hz}$
Alternately, $\frac{f_a}{f_r} = \frac{v + v_s}{v - v_s} = \frac{340 + 20}{340 - 20} = \frac{9}{8}$
$\Rightarrow f_a : f_r = 9 : 8$