A bat flies at a steady speed of `4 ms^(-1)` emitting 90 kHz sound waves and is flying towards a wall. It detects a reflected signal at a frequency (speed of sound is `340ms^(-1)`)

Options

A. 90.2 kHz
B. 91.2 kHz
C. 92.2 kHz
D. 93.2 kHz

Explanation

To solve the problem, we need to determine the frequency of the sound waves detected by the bat after they reflect off the wall. We will use the Doppler effect formula for sound waves.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Speed of the bat (observer), \( v_o = 4 \, \text{m/s} \)
- Frequency emitted by the bat (source), \( f_0 = 90 \, \text{kHz} = 90 \times 10^3 \, \text{Hz} \)
- Speed of sound in air, \( v = 340 \, \text{m/s} \)

2. **Understand the Situation:**
- The bat is moving towards a wall, emitting sound waves. When the sound waves hit the wall, they reflect back towards the bat. The bat will detect these reflected waves at a different frequency due to the Doppler effect.

3. **Use the Doppler Effect Formula:**
The formula for the apparent frequency \( f' \) when both the source and observer are moving towards each other is given by:
\[ f' = f_0 \frac{v + v_o}{v - v_s} \]
where:
- \( f' \) = apparent frequency
- \( f_0 \) = original frequency
- \( v \) = speed of sound
- \( v_o \) = speed of the observer (bat)
- \( v_s \) = speed of the source (the wall is stationary, so \( v_s = 0 \))

4. **Calculate the Reflected Frequency:**
- First, we need to calculate the frequency of the sound waves as they reach the wall (the wall acts as a stationary source):
\[ f' = f_0 \frac{v + v_o}{v} = 90 \times 10^3 \frac{340 + 4}{340} \]
\[ f' = 90 \times 10^3 \frac{344}{340} \]
\[ f' = 90 \times 10^3 \times 1.01176 \approx 90 \times 10^3 \times 1.01176 \approx 91058.4 \, \text{Hz} \]

5. **Calculate the Frequency Detected by the Bat:**
Now, the reflected sound waves will act as a source moving towards the bat. The frequency detected by the bat can be calculated using the same formula:
\[ f'' = f' \frac{v + v_o}{v} \]
where \( f' \) is the frequency we just calculated.
\[ f'' = 91058.4 \frac{340 + 4}{340} \]
\[ f'' = 91058.4 \times 1.01176 \approx 92000 \, \text{Hz} \]

6. **Convert to kHz:**
\[ f'' \approx 92 \, \text{kHz} \]

### Final Answer:
The frequency detected by the bat after reflecting off the wall is approximately **92 kHz**.