Question

A person at rest hears an electric siren which is stationary. Now the person accelerates at \(2\,\text{m s}^{-2}\) along a straight line path. The distance travelled by him when he hears the frequency of the siren as \(94\%\) of its original value is \((\text{speed of sound} = 330\,\text{m s}^{-1})\):

• A. \(49\,\text{m}\)
• B. \(98\,\text{m}\)
• C. \(147\,\text{m}\)
• D. \(196\,\text{m}\)

Hint:

For Doppler effect involving moving observer: \[ f'=f\left(\frac{v\pm v_o}{v}\right) \] Use minus sign when observer moves away from source.

Answer 1

The Correct Option is B

Step 1: Set up the Doppler case.
The siren stays still and the listener moves. When the listener moves away from a still source, the heard frequency drops. The formula is \[ f' = f\left(\frac{v - v_o}{v}\right) \] where $v$ is sound speed and $v_o$ is the listener's speed.

Step 2: Use the frequency condition.
The heard frequency is $94\%$ of the real one, so $f' = 0.94 f$. \[ 0.94 = \frac{330 - v_o}{330} \]
Step 3: Solve for the listener's speed.
Multiply both sides by $330$. \[ 310.2 = 330 - v_o \] \[ v_o = 19.8 \approx 20\,\text{m/s} \]
Step 4: Switch to the motion of the person.
The person starts from rest and speeds up at $a = 2$ m/s$^2$. We need the distance covered to reach $20$ m/s.
Step 5: Use the speed distance equation.
\[ v^2 = u^2 + 2 a s \] \[ 20^2 = 0 + 2(2)s \]
Step 6: Solve for the distance.
\[ 400 = 4s \quad\Rightarrow\quad s = 100\,\text{m} \] The nearest given choice is \[ \boxed{98\,\text{m}} \]