Question

A car is initially at rest at a distance of $330\text{m}$ from a stationary observer. It begins to move towards the observer with an acceleration of $1.1{\text{m/s}}^2$, sounding its horn continuously. $20\text{s}$ later the driver stops sounding the horn. The observer will hear the sound of horn for a duration of:
(Take velocity of sound in air to be $330\text{m/s})$

• A. $20\text{s}$
• B. $\frac{58}{3}\text{s}$
• C. $\frac{61}{3}\text{s}$
• D. $21\text{s}$

Answer 1

The correct option is B $\frac{58}{3}\text{s}$


Let at time $t=0\text{s}$, car begins to move towards observer at point $\text{B}$.

The sound wave emitted at $t=0$ from the source at position $\text{A}$, reaches the observer at $\text{B}$ after time,

$\Delta t_1=\frac{330\text{m}}{330\text{m/s}}=1\text{s}$

Hence observer at $\text{B}$ started hearing the sound at $t=1\text{s}$.

After $20\text{s}$ distance travelled by the source is,

$S=\frac{1}{2}a{t}^2$

$\Rightarrow S=\frac{1}{2}\times 1.1\times (20{)}^2=220\text{m}$

Hence at $t=20\text{s}$ the source reaches at point $\text{C}$ i.e. at $220\text{m}$ from $\text{A}$.

Since the driver stops sounding the horn after $20\text{s}$, the last sound wave emitted by the source at $\text{C}$ will reach the observer at $\text{B}$ after

$\Delta t_2=\frac{d}{v_{\text{sound}}}=\frac{110\text{m}}{330\text{m/s}}$

$\Rightarrow \Delta t_2=\frac{1}{3}\text{s}$

Thus the observer will not hear the sound for entire $20\text{s}$. Instead the duration of sound heard by the observer is

$\Delta t=(20-\Delta t_1)+\Delta t_2$

$\Rightarrow \Delta t=(20-1)+\frac{1}{3}$

$\therefore \Delta t=19+\frac{1}{3}=\frac{58}{3}\text{s}$

$\begin{array}{c}\text{Why this question}?\\ \text{Caution: Always remember that once the sound pulse is}\\ \text{emitted from the source, it is no longer a part of}\\ \text{the source and will propagate with the speed of sound}\\ \text{in the given medium.}\end{array}$