Question

A radar sends sound waves of frequency 'f' towards an aeroplane moving away from it. If the difference in the frequencies of waves sent and received by the radar is 10% of 'f', then the speed of the aeroplane in $ms^{-1}$ is nearly (Speed of sound in air = 342 $ms^{-1}$)

• A. 15
• B. 33
• C. 30
• D. 18

Hint:

For reflection from a moving object, the Doppler shift formula for $\Delta f$ when $v \ll c$ is approx $\Delta f = \frac{2v}{c}f$. Using approximation: $0.1 = \frac{2v}{342} \Rightarrow v = 17.1$, which is close to 18. Exact calculation is safer here.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a Doppler effect problem involving sound waves where the source (radar) is stationary and the "observer" (plane) acts as a moving reflector. 1. The plane receives frequency $f'$ (Moving observer). 2. The plane reflects this frequency as a moving source, sending $f''$ back to the radar.
Step 2: Formula Derivation:
Let $c$ be the speed of sound and $v$ be the speed of the plane moving away. Frequency received by plane ($f'$): \[ f' = f \left( \frac{c - v}{c} \right) \] Frequency received back by radar ($f''$) (Plane acts as source moving away): \[ f'' = f' \left( \frac{c}{c + v} \right) = f \left( \frac{c - v}{c} \right) \left( \frac{c}{c + v} \right) = f \left( \frac{c - v}{c + v} \right) \]
Step 3: Calculation:
Given frequency difference $\Delta f = f - f'' = 10%$ of $f = 0.1f$. \[ f - f \left( \frac{c - v}{c + v} \right) = 0.1f \] \[ 1 - \frac{c - v}{c + v} = 0.1 \] \[ \frac{(c + v) - (c - v)}{c + v} = 0.1 \] \[ \frac{2v}{c + v} = 0.1 = \frac{1}{10} \] \[ 20v = c + v \] \[ 19v = c \] Given $c = 342 \, ms^{-1}$. \[ v = \frac{342}{19} \] \[ v = 18 \, ms^{-1} \]
Step 4: Final Answer:
The speed of the aeroplane is 18 $ms^{-1}$.