Question

A bat emitting an ultrasonic wave of frequency \( 4.5 \times 10^4 \, \text{Hz} \) at speed of \( 6 \, \text{m/s} \) between two parallel walls. The two frequencies heard by the bat will be

• A. \( 4.67 \times 10^4 \, \text{Hz}, 4.34 \times 10^4 \, \text{Hz} \)
• B. \( 4.34 \times 10^4 \, \text{Hz}, 4.67 \times 10^4 \, \text{Hz} \)
• C. \( 4.5 \times 10^4 \, \text{Hz}, 5.4 \times 10^4 \, \text{Hz} \)
• D. \( 4.67 \times 10^3 \, \text{Hz}, 4.34 \times 10^4 \, \text{Hz} \)

Hint:

In Doppler effect problems, remember to adjust the frequency based on whether the source is moving towards or away from the observer.

Answer 1

The Correct Option is A

The phenomenon described is the Doppler effect, where the frequency of the sound heard by the bat will be altered due to the motion of the bat relative to the walls. - The frequency heard by the bat when the sound is approaching the wall is increased. This is given by: \[ f' = f \left( \frac{v + v_{\text{source}}}{v} \right) \] where: - \( f = 4.5 \times 10^4 \, \text{Hz} \),
- \( v = 6 \, \text{m/s} \) (the speed of sound),
- \( v_{\text{source}} = 6 \, \text{m/s} \) (the speed of the bat). Thus, the frequency heard by the bat when the sound is approaching the wall is: \[ f' = 4.5 \times 10^4 \times \left( \frac{6 + 6}{6} \right) = 4.67 \times 10^4 \, \text{Hz} \] - The frequency heard by the bat when the sound is reflected from the wall is decreased. This is given by: \[ f'' = f \left( \frac{v - v_{\text{source}}}{v} \right) \] Thus: \[ f'' = 4.5 \times 10^4 \times \left( \frac{6 - 6}{6} \right) = 4.34 \times 10^4 \, \text{Hz} \] Therefore, the two frequencies heard by the bat are \( 4.67 \times 10^4 \, \text{Hz} \) and \( 4.34 \times 10^4 \, \text{Hz} \), so the correct answer is (A).