Question

A highway truck has two horns $A$ and $B$. When sounded together, the driver records $50$ beats in $10$ seconds. With horn $B$ blowing and the truck moving towards a wall at a speed of $10m/s$, the driver noticed a beat frequency of $5Hz$ with the echo. When frequency of $A$ is decreased the beat frequency with two horns sounded together increases. Calculate the frequency of horn $A$. $($ Speed of sound in air $=330m/s)$

• A. 75 HZ
• B. 85 Hz
• C. 90 HZ
• D. 95 Hz

Answer 1

Answer: (A)

Here, let frequencies of horn $A$ and $B$ are $n_A$ and $n_B$, respectively.
As given, $n_A-n_B=\frac{50}{10}$
$n_A-n_B=5$ beats
When, the horn $B$ is blown while moving towards the wall the apparent frequency,
$n_B^{{}^{\prime }}=n_B\frac{v+v_0}{v-v_s}$
Since, both the observer and source are moving with truck.
So $v_s=v_0=10m/s$
$n_B^{{}^{\prime }}=n_B\left(\frac{330+10}{330-10}\right)$
$n_B^{{}^{\prime }}=n_{B}1.0625$
$=n_B{}^{{}^{\prime }}-n_B=0.0625n_B$
As given, $n_B^{{}^{\prime }}-n_B=5$
$0.0625n_B=5$
$n_B=80Hz$
$\because$ Given, that if frequency of $A$ is decreased then beat frequency is increases.
So, $n_B>n_A$
$\therefore n_A=n_B-5=80-5$
$\Rightarrow n_A=75Hz$