The velocity of sound is $340 \mathrm{~m} / \mathrm{s}$. A source of sound having frequency of $90…
The velocity of sound is $340 \mathrm{~m} / \mathrm{s}$. A source of sound having frequency of $90 \mathrm{~Hz}$ is moving towards a stationary observer with a speed of one-tenth that of sound. The apparent frequency of sound as heard by the observer is
$45 \mathrm{~Hz}$
$100 \mathrm{~Hz}$
$80 \mathrm{~Hz}$
$50 \mathrm{~Hz}$
Solution
Apparent frequency of sound heard by the observer is:
$f=f_0\left(\frac{v}{v-v_s}\right)$
Where, $v=340 \mathrm{~m} / \mathrm{s}$ is the speed of sound \& $v_s=v / 10$ is the speed of the source morning towards the observer.
$\therefore f=90\left(\frac{v}{v-\frac{v}{10}}\right) \mathrm{Hz}=90\left(\frac{10}{9}\right) \mathrm{Hz}=100 \mathrm{~Hz}$