79. A bat emitting ultrasound at 50 kHz is flying directly towards a solid wall with a speed of 3 ms–1. If the speed of sound in air is 330 ms–1, the frequency of the reflected signal (in kHz) heard by the bat will be
Bat and Wall Doppler Effect – Finding the Reflected Ultrasound Frequency Heard by the Bat
Given Data
Frequency emitted by bat, f = 50 kHz
Speed of bat, vb = 3 m/s
Speed of sound, v = 330 m/s
Wall is stationary.
Concept Used
Since the wall is stationary, the problem involves two successive Doppler shifts:
Step 1: The moving bat acts as the source and the stationary wall acts as the observer.
Step 2: The reflected wave now originates from the stationary wall (acting as a new source), while the moving bat acts as the observer.
Therefore, Doppler Effect is applied two times.
Step 1: Frequency Received by the Wall
When the source moves towards a stationary observer, the observed frequency is
f₁ = f × v / (v − vb)
Substituting the values,
f₁ = 50 × 330 / (330 − 3)
= 50 × 330 / 327
≈ 50.46 kHz
This is the frequency with which the stationary wall receives the sound wave.
Step 2: Frequency Heard by the Bat After Reflection
The wall reflects the sound without changing its frequency. Therefore, the reflected wave has frequency 50.46 kHz.
Now the wall acts as a stationary source, while the bat moves towards it.
The observed frequency is
f₂ = f₁ × (v + vb) / v
Substituting the values,
f₂ = 50.46 × (330 + 3) / 330
= 50.46 × 333 / 330
≈ 50.92 kHz
Shortcut Formula
Instead of applying the Doppler Effect twice, we can directly use the combined formula for reflection from a stationary wall:
f′ = f × (v + u) / (v − u)
where
- f = original frequency
- u = speed of the bat
- v = speed of sound
Substituting the values,
f′ = 50 × (330 + 3) / (330 − 3)
= 50 × 333 / 327
= 50 × 1.01835
≈ 50.92 kHz
Why Does the Frequency Increase?
The bat is flying towards the wall, so the sound waves become compressed before reaching the wall. As a result, the wall receives a frequency higher than the emitted frequency.
After reflection, the bat is still moving towards the incoming reflected waves. Because the observer is moving towards the source, the bat encounters the wavefronts more frequently, causing another increase in the observed frequency.
Thus, the frequency increases twice—once during emission and once during reception—making the reflected frequency noticeably higher than the original frequency.
Detailed Explanation of the Physics
Reflection itself does not change the frequency of a sound wave when the reflecting surface is stationary. The wall simply sends the same frequency back into the air. The change in frequency occurs only because of the relative motion between the bat and the sound wave before reflection and again after reflection.
If the wall were also moving, the analysis would become more complex because the wall would no longer behave as a stationary observer or source. However, since the wall is fixed, the two-step Doppler approach gives the correct answer.
Final Answer
Frequency of the reflected ultrasound heard by the bat = 50.92 kHz
Answer: 50.9 kHz (approximately)
