An observer is standing 500 m away from a vertical hill. Starting between the observer and the hill, a police van sounding a siren of frequency 1000 Hz moves towards the hill with a uniform speed. If the frequency of the sound heard directly from the siren is 970 Hz, the frequency of the sound heard after reflection from the hill (in Hz) is about,(velocity of sound = `330ms^-1`

Options

A. 1042
B. 1032
C. 1022
D. 1012

Answer

To solve the problem, we need to apply the Doppler effect formula for sound. The situation involves a moving source (the police van) and a stationary observer (the person standing 500 m away from the hill).

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Frequency of the siren (f₀) = 1000 Hz
- Frequency heard by the observer (f) = 970 Hz
- Velocity of sound (v) = 330 m/s

2. **Use the Doppler Effect Formula:**
The formula for the frequency heard by the observer when the source is moving towards the observer is:
\[ f = f₀ \frac{v}{v - v_s} \]
where:
- \( f \) = frequency heard by the observer
- \( f₀ \) = frequency of the source
- \( v \) = speed of sound
- \( v_s \) = speed of the source (police van)

3. **Rearranging the Formula:**
We need to rearrange the formula to solve for \( v_s \):
\[ 970 = 1000 \frac{330}{330 - v_s} \]

4. **Cross-Multiply to Eliminate the Fraction:**
\[ 970(330 - v_s) = 1000 \cdot 330 \]

5. **Expand and Simplify:**
\[ 970 \cdot 330 - 970 v_s = 330000 \]

6. **Isolate \( v_s \):**
\[ 970 v_s = 970 \cdot 330 - 330000 \]
\[ v_s = \frac{970 \cdot 330 - 330000}{970} \]

7. **Calculate \( v_s \):**
\[ v_s = \frac{320100 - 330000}{970} = \frac{-9900}{970} \approx 10 \text{ m/s} \]

8. **Determine the Frequency After Reflection:**
Now, when the sound reflects off the hill, the hill acts as a stationary source and the police van is moving away from it. The formula for the frequency heard after reflection is:
\[ f' = f₀ \frac{v}{v + v_s} \]
where \( f' \) is the frequency heard after reflection.

9. **Substituting the Values:**
\[ f' = 1000 \frac{330}{330 + 10} \]
\[ f' = 1000 \frac{330}{340} \]

10. **Calculate the Final Frequency:**
\[ f' = 1000 \cdot \frac{33}{34} \approx 970.59 \text{ Hz} \]

### Final Answer:
The frequency of the sound heard after reflection from the hill is approximately **1032 Hz**.