A source emits sound with a frequency of 1000 Hz. It is moving at 20 m/s toward a stationary reflecting wall. If the speed of sound is 340 m/s an observer at rest directly behind the source hears a beat frequency of

Options

A. 11 Hz
B. 86 Hz
C. 97 Hz
D. 118 Hz

Answer

To solve the problem, we need to determine the beat frequency heard by an observer behind a moving sound source. The sound source emits a frequency of 1000 Hz and is moving towards a stationary reflecting wall at a speed of 20 m/s. The speed of sound is given as 340 m/s.

### Step-by-Step Solution:

1. **Identify the Frequencies**:
- The source emits a sound with a frequency \( f_s = 1000 \, \text{Hz} \).
- The speed of sound in air is \( v = 340 \, \text{m/s} \).
- The speed of the source is \( v_s = 20 \, \text{m/s} \).

2. **Calculate the Frequency Heard by the Reflecting Wall**:
- The wall acts as a stationary observer. The frequency \( f_1 \) heard by the wall can be calculated using the Doppler effect formula:
\[ f_1 = f_s \frac{v}{v - v_s} \]
- Plugging in the values:
\[ f_1 = 1000 \, \text{Hz} \times \frac{340 \, \text{m/s}}{340 \, \text{m/s} - 20 \, \text{m/s}} = 1000 \, \text{Hz} \times \frac{340}{320} \]
- Simplifying this:
\[ f_1 = 1000 \, \text{Hz} \times 1.0625 = 1062.5 \, \text{Hz} \]

3. **Calculate the Frequency Heard by the Observer**:
- The observer is stationary and hears the sound directly from the source. The frequency \( f_2 \) heard by the observer can be calculated as:
\[ f_2 = f_s \frac{v + 0}{v + v_s} \]
- Since the observer is stationary, the speed of the observer \( v_o = 0 \):
\[ f_2 = 1000 \, \text{Hz} \times \frac{340 \, \text{m/s}}{340 \, \text{m/s} + 20 \, \text{m/s}} = 1000 \, \text{Hz} \times \frac{340}{360} \]
- Simplifying this:
\[ f_2 = 1000 \, \text{Hz} \times 0.9444 \approx 944.4 \, \text{Hz} \]

4. **Calculate the Beat Frequency**:
- The beat frequency \( f_b \) is the difference between the two frequencies:
\[ f_b = |f_1 - f_2| = |1062.5 \, \text{Hz} - 944.4 \, \text{Hz}| \]
- Calculating this gives:
\[ f_b = 118.1 \, \text{Hz} \approx 118 \, \text{Hz} \]

### Final Answer:
The beat frequency heard by the observer is approximately **118 Hz**.