Q52E - University Physics with Modern Physics

Question

While sitting in your car by the side of a country road, you are approached by your friend, who happens to be in an identical car. You blow your car’s horn, which has a frequency of 260 Hz. Your friend blows his car’s horn, which is identical to yours, and you hear a beat frequency of 6.0 Hz. How fast is your friend approaching you?

Step-by-Step Solution

Answer

Friends approaching at the rate of $-7.76m/s$

Step 1:

Given data:

As by the doppler effect:

$f_L=\frac{v+v_L}{v+v_S}{f}_S$

$f_L=$Frequency observed by the listener

$v=$speed of sound

$v_L=$speed of listener

$f_s=$frequency of source

$v_s=$speed of the source of sound

And here $v_L$ is positive as the velocity of the listener from Listener to Source and $v_s$ is positive as the velocity of the source from Listener to Source and velocity is negative.

The frequency of the sound heard by the listener is not the same as the source frequency when the source and listener are moving relative to each other.

Step 2:

The moving car is the source of the sound that is shifted by the doppler effect.

$$\begin{gathered} f_{\mathrm{s}}\to \mathrm{The} \mathrm{moving} \mathrm{car} \\ {\mathrm{f}}_{\mathrm{beat}}={\mathrm{f}}_{\mathrm{L}}-{\mathrm{f}}_{\mathrm{s}} \\ {\mathrm{f}}_{\mathrm{L}}={\mathrm{f}}_{\mathrm{beat}}+{\mathrm{f}}_{\mathrm{s}}=260+6=266\mathrm{Hz} \\ {\mathrm{v}}_{\mathrm{L}}=0 \end{gathered}$$

Here, $v_s\to$(-) as the source is moving towards the listener hence it is negative.

$$\begin{gathered} f_L=\frac{v+v_L}{v+v_s}{f}_s \\ v+v_s=f_s\frac{v+v_L}{f_L} \end{gathered}$$

On putting the values;

$$\begin{gathered} v_s=f_s\frac{v+v_L}{f_L}-v \\ =\left(260Hz\right)\frac{344m/s}{266Hz}-344m/s \\ =-7.76m/s \end{gathered}$$

Hence, friends approaching at the rate of $-7.76m/s$