Question

The source $S_1$ is at rest. The observer and the source $S_2$ are moving towards $S_1$ as shown in the figure. The roof beats observed by the observer if both sources have a frequency of 120 Hz and a speed of sound of $330m{s}^{-1}$ is:

Answer 1

Step 1: Given data:-

There are two sources $S_1$ and $S_2$ and $S_2$ is moving towards $S_1$ which is at rest.

$$\begin{gathered} {\mathrm{f}}_{{\mathrm{S}}_1}={\mathrm{f}}_{{\mathrm{S}}_2}=120\mathrm{Hz} \\ {\mathrm{v}}_{{\mathrm{s}}_2}=108\mathrm{Km}/\mathrm{m}=30{\mathrm{ms}}^{-1} \\ {\mathrm{v}}_{\mathrm{sound}}=330{\mathrm{ms}}^{-1} \\ {\mathrm{v}}_{\mathrm{observer}}=10{\mathrm{ms}}^{-1} \\ {\mathrm{v}}_{\mathrm{observer}}{\mathrm{cos\theta }}_{\mathrm{o}}=10\cos {53}^{\mathrm{o}} \\ {\mathrm{v}}_{{\mathrm{s}}_2}{\mathrm{cos\theta }}_{{\mathrm{s}}_2}=30\cos {37}^{\mathrm{o}} \end{gathered}$$

Step: 2 Formula used,

$f_o=\left(\frac{(𝝂+{𝝂}_s)}{(𝝂-{𝝂}_o)}\right)f_s$

Here,$$f_o$$ = observer frequency of sound

$$f_s$$ = actual frequency of sound waves

v = speed of sound waves

$$v_o$$ = observer velocity

$$v_s$$ =source velocity

Here the roof is the observer,

$f_{roof}$is the frequency of the observer.

${\mathrm{f}}_{\mathrm{roof}}={\mathrm{f}}_{{\mathrm{s}}_2}\left\{\frac{{\mathrm{v}}_{\mathrm{sound}}+{\mathrm{v}}_{\mathrm{observer}}{\mathrm{cos\theta }}_{\mathrm{o}}}{{\mathrm{v}}_{\mathrm{sound}}-{\mathrm{v}}_{{\mathrm{s}}_2}{\mathrm{cos\theta }}_{{\mathrm{s}}_2}}\right\}-\left\{\frac{{\mathrm{v}}_{\mathrm{sound}}+{\mathrm{v}}_{\mathrm{observer}}}{{\mathrm{v}}_{\mathrm{sound}}}\right\}$

$f_{s_2}$ is the frequency of the source

Step: 3 Solving for ${\mathrm{f}}_{\mathrm{roof}}$

In the case of the opposite when an object is going far from each other, we got

$$\begin{gathered} v_{observer}=10\cos {53}^o \\ {v}_{s_2}=30\cos {37}^o \end{gathered}$$

When both objects are coming close to each other then the apparent frequency will increase. So, we will take positive in the numerator and negative in the denominator.

In this case, $v_{s_2}=0$ because the source is stationary,

$$\begin{gathered} {\mathrm{f}}_{\mathrm{roof}}=120\left\{\left\{\frac{330+10\cos {53}^{\mathrm{o}}}{330-30\cos {37}^{\mathrm{o}}}\right\}-\left\{\frac{330+10}{330}\right\}\right\} \\ {\mathrm{f}}_{\mathrm{roof}}=120\left\{\left\{\frac{330+8.337}{330-24}\right\}-\left\{\frac{340}{330}\right\}\right\} \\ {\mathrm{f}}_{\mathrm{roof}}=120\left\{\left\{\frac{338.337}{306}\right\}-\left\{\frac{340}{330}\right\}\right\} \\ {\mathrm{f}}_{\mathrm{roof}}=120\left\{\left\{1.105\right\}-\left\{1.03\right\}\right\} \\ {\mathrm{f}}_{\mathrm{roof}}=120\left\{0.0746\right\} \\ {\mathrm{f}}_{\mathrm{roof}}=8.96\mathrm{Hz} \end{gathered}$$

Therefore, the roof beats observed by the observer is 8.96Hz.