Question

A supersonic jet plane moves parallel to the ground at speed $v=0.75$ Mach (1 Mach=speed of sound). The frequency of its engine sound is ${{v}_{0}}=2KHz$ and the height of the jet plane is $h=1.5km$. At some instant an observer on the ground hears a sound of frequency ${{v}_{0}}={{2}_{0}}$. Find the instant (in second) prior to the instant of hearing when the sound wave received by the observer was emitted by the jet plane. Velocity of sound wave in the condition is observer $=340m/s$

Answer 1

Hint:When there is relative motion between the observer and the source of the sound then to find the frequency of the sound observed by the observer we use Doppler’s effect formula. To find the time taken to cover a distance with constant speed we use the motion formula.


Complete step-by-step solution:
When there is relative motion between the source of the sound and the observer then the apparent frequency of the sound observed by the observer is less or greater than the original frequency based on the direction of the motion of the source of the sound.
If the source of sound is moving away from the observer then the apparent frequency observed is less than the original frequency of the sound.
If the source of sound is moving towards the observer then the apparent frequency observed is greater than the original frequency of the sound.
Using Doppler’s effect formula,
${{f}_{app}}={{f}_{o}}\left( \dfrac{{{v}_{sound}}-{{v}_{observer}}}{{{v}_{sound}}-{{v}_{source}}} \right)$
It is given that the observed frequency is twice the original frequency of the sound and the observer is at rest.
$2{{f}_{o}}={{f}_{o}}\left( \dfrac{{{v}_{sound}}-0}{{{v}_{sound}}-{{v}_{jet}}\cos \theta } \right)$
Here, $\theta $ is the angle made by the direction of the jet with the horizontal.
\[\begin{align}
  & \Rightarrow 2=\left( \dfrac{{{v}_{sound}}}{{{v}_{sound}}-0.75{{v}_{sound}}\cos \theta } \right) \\
 & \Rightarrow 2=\dfrac{1}{1-0.75\cos \theta } \\
 & \Rightarrow 2-1.5\cos \theta =1 \\
 & \Rightarrow 1.5\cos \theta =1 \\
 & \Rightarrow \cos \theta =\dfrac{1}{1.5} \\
 & \Rightarrow \cos \theta =\dfrac{2}{3}
\end{align}\]
Using trigonometric identity,
$\begin{align}
  & \sin \theta =\sqrt{1-{{\cos }^{2}}\theta } \\
 & =\sqrt{1-{{\left( \dfrac{2}{3} \right)}^{2}}} \\
 & =\sqrt{1-\dfrac{4}{9}} \\
 & \sin \theta =\dfrac{\sqrt{5}}{3}
\end{align}$
The distance travelled by the jet in the direction perpendicular to the horizontal$=d=\dfrac{h}{\sin \theta }$
$d=\dfrac{1500}{\left( \dfrac{\sqrt{5}}{3} \right)}m=2012.5m$
Time taken by the sound $t=\dfrac{dis\tan ce}{speed}=\dfrac{2012.5}{340} \approx 6\sec $
Hence, the time taken by sound to travel the distance is approximately equal to 6 sec.

Note:For observed value of frequency we use Doppler’s effect formula.
The relative velocity of the source of the sound is calculated in the direction of line joining the source of the sound and the observer.