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(D) According to the Doppler effect,when a source moves towards a stationary observer,the observed frequency $f$ is given by $f = f_0 \left( \frac{v}{

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$\frac{19}{18}$

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(D) According to the Doppler effect,when a source moves towards a stationary observer,the observed frequency $f$ is given by $f = f_0 \left( \frac{v}{v - v_s} ight)$,where $f_0$ is the source frequency,$v$ is the speed of sound,and $v_s$ is the speed of the source.For the first case,$v_s = 34 \, m/s$:$f_1 = f_0 \left( \frac{340}{340 - 34} ight) = f_0 \left( \frac{340}{306} ight) = f_0 \left( \frac{10}{9} ight)$.For the second case,$v_s = 17 \, m/s$:$f_2 = f_0 \left( \frac{340}{340 - 17} ight) = f_0 \left( \frac{340}{323} ight) = f_0 \left( \frac{20}{19} ight)$.Now,calculating the ratio $\frac{f_1}{f_2}$:$\frac{f_1}{f_2} = \frac{f_0 \left( \frac{10}{9} ight)}{f_0 \left( \frac{20}{19} ight)} = \frac{10}{9} 	imes \frac{19}{20} = \frac{19}{18}$.

Similar Questions

$A$ boy is walking away from a wall towards an observer at a speed of $1\, m/s$ and blows a whistle whose frequency is $680\, Hz$. The number of beats heard by the observer per second is (Velocity of sound in air $= 340\, m/s$).

With what velocity an observer should move relative to a stationary source so that a sound of double the frequency of the source is heard by the observer?

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