Question

A band playing music at frequency $f$ is moving towards a wall at a speed $v_b$. A motorist is following the band with a speed $v_m$. If $v$ is the speed of sound, the expression for the beat frequency heard by the motorist is

• A. $\frac{V+V_m}{V+V_b}f$
• B. $\frac{V+V_m}{V-V_b}f$
• C. $\frac{2V_b\left(V+V_m\right)}{V^2-V_b^2}f$
• D. $\frac{2V_m\left(V+V_b\right)}{V^2-V_m^2}f$

Answer 1

Answer: (C)

The motorist receives two sound waves are direct one and that reflected from the wall.
$f^{\prime }=\left(\frac{v+v_m}{v+v_b}\right)f$
For reflected sound waves:
Frequency of sound wave reflected from the wall is,
$f^{\prime \prime }=\frac{v}{v-v_b}\times f$
Frequency of the reflected waves as received by the moving motorist is
$f^{\prime \prime \prime }=\frac{v+v_m}{v}\times f^{\prime \prime }=\frac{v+v_m}{v-v_b}\times f$
Therefore, the beat frequency is
$f^{\prime \prime \prime }-f^{\prime }=\frac{v+v_m}{v-v_b}\times f-\frac{v+v_m}{v+v_b}\times f$
$=\frac{2v_b\left(v+v_m\right)}{\left(v^2-v_b^2\right)}f$