Question

The driver of a bus approaching a big wall notices that the frequency of his bus’s horn changes from $420Hzto490Hz$ when he hears it after it gets reflected from the wall. Find the speed of the bus if the speed of the sound is $330m{s}^{-1}$.

• A. $81km/h$
• B. $91km/h$
• C. $71km/h$
• D. $61km/h$

Answer 1

The correct option is B

$91km/h$

Step 1: Given data

Initial frequency$\left(f_{\circ }\right)=420Hz$

Increased frequency$\left(f\right)=490Hz$

Speed of sound $\left(c\right)=330m/s$

Step 2: Formula used

Using the Doppler effect formula, the frequency can be calculated as

$f\text{'}=\frac{v+v_o}{v-v_s}{f}_o\dots \dots \dots \dots \dots \dots \left(1\right)$

where $f\text{'}=$Observed frequency, $f_o=$the actual frequency of the sound wave, $v_o=$velocity of the observer, $v_s=$speed of the source relative to the medium, $v=\left(c\right)=330m/s$ speed of sound

Step 3: Compute the speed of the bus

Consider the following figure that represents the situation before reflection where the changed frequency is $f_1$and the velocity of the observer which is a wall is zero and the actual frequency of the bus's horn and speed of the source is the same as the bus's speed.

Let the bus speed is $v_b$

From Doppler's effect,

$\begin{array}{rcl}f\text{'}& =& \frac{v+v_o}{v-v_s}{f}_o\\ f_1& =& \frac{v+0}{v-v_b}\times f_o\\ f_1& =& \frac{v}{v-v_b}\times 420\\ f_1& =& \frac{330}{330-v_b}\times 420\dots \dots \dots \dots \dots \dots \dots \left(2\right)\end{array}$

And the situation after reflection is given in the following figure

Now after reflection $f_1$be the actual frequency and $f_2$ be the changed frequency and the observer speed will be the speed of the bus that is $v_o=v_b$ and the speed of the source which is the wall is zero that is $v_s=0$

From Doppler's effect

$$\begin{gathered} f_2=\frac{v+v_b}{v-0}\times f_1 \\ {f}_2=\frac{330+v_b}{v}\times f_1 \end{gathered}$$

Now from equation (2) substituting the values of $f_1$

$$\begin{gathered} f_2=\frac{330+v_b}{330}\times \frac{330}{330-v_b}\times 420 \\ 490=\frac{330+v_b}{330-v_b}\times 420 \\ \frac{7}{6}=\frac{330+v_b}{330-v_b} \end{gathered}$$

Now using componenedo-dividendo rule

$$\begin{gathered} \frac{7+6}{7-6}=\frac{330+v_b+330-v_b}{330+v_b-330+v_b} \\ \frac{13}{1}=\frac{660}{2v_b} \\ {v}_b=\frac{660}{2\times 13} \\ {v}_b=\frac{330}{13}\mathrm{m}/\mathrm{s}=\frac{330}{13}\times \frac{18}{5}=91.38\approx 91\mathrm{km}/\mathrm{h} \end{gathered}$$

Thus, the speed of the bus is $91km/hr$.

Hence, option B is the correct answer.