a bus is moving towards and stationary observer with speed 20 m/s blows a horn of frequency 200 Hz. the frequency of sound heard by the observer is approximately ( take speed of sound 320 m/s)

Options

A. 188 Hz
B. 200 Hz
C. 226 Hz
D. 213 Hz

Answer

To solve the problem, we will use the Doppler effect formula for sound. The formula for the frequency heard by a stationary observer when the source of sound is moving towards the observer is given by:

\[ f' = \frac{v + v_o}{v - v_s} f_0 \]

Where:
- \( f' \) = frequency heard by the observer
- \( f_0 \) = frequency of the source (200 Hz)
- \( v \) = speed of sound in air (320 m/s)
- \( v_o \) = speed of the observer (0 m/s, since the observer is stationary)
- \( v_s \) = speed of the source (20 m/s, since the bus is moving towards the observer)

### Step-by-Step Solution:

1. **Identify the values**:
- \( f_0 = 200 \, \text{Hz} \)
- \( v = 320 \, \text{m/s} \)
- \( v_o = 0 \, \text{m/s} \) (observer is stationary)
- \( v_s = 20 \, \text{m/s} \) (bus is moving towards the observer)

2. **Substitute the values into the formula**:
\[ f' = \frac{320 + 0}{320 - 20} \times 200 \]

3. **Simplify the expression**:
\[ f' = \frac{320}{300} \times 200 \]

4. **Calculate \( \frac{320}{300} \)**:
\[ \frac{320}{300} = \frac{32}{30} = \frac{16}{15} \]

5. **Now multiply by 200**:
\[ f' = \frac{16}{15} \times 200 = \frac{3200}{15} \]

6. **Perform the division**:
\[ f' = 213.33 \, \text{Hz} \]

Thus, the frequency of sound heard by the observer is approximately **213.33 Hz**.

### Final Answer:
The frequency of sound heard by the observer is approximately **213 Hz**.

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