Two cars are approaching each other at an equal speed of 7.2 km / hr.

Question:

Two cars are approaching each other at an equal speed of $7.2 \mathrm{~km} / \mathrm{hr}$. When they see each other, both blow horns having frequency of $676 \mathrm{Hz.}$. The beat frequency heard by each driver will be_____ Hz. [Velocity of sound in air is $340 \mathrm{~m} / \mathrm{s}$.]

Solution:

Speed $=7.2 \mathrm{~km} / \mathrm{h}=2 \mathrm{~m} / \mathrm{s}$

Frequency as heard by A $f_{A}^{\prime}=f_{B}\left(\frac{V+V_{0}}{V-V_{n}}\right)$

$f_{A}^{\prime}=676\left(\frac{340+2}{340-2}\right)$

$f_{A}^{\prime}=684 H z$

$\therefore \mathrm{f}_{\text {Beat }}=\mathrm{f}_{\mathrm{A}}^{\prime}-\mathrm{f}_{\mathrm{B}}$

$=684-676$

$=8 \mathrm{~Hz}$

Leave a comment