A policeman on duty detects a drop of 10% in the pitch of the hom of a motor car as it crosses him. If the velocity of sound is 330 m/s calculate the speed of car?

Answer

To solve the problem, we will use the Doppler effect equations for sound. The key points to remember are that the frequency detected by the observer changes based on the relative motion between the source (the car) and the observer (the policeman).

### Step-by-Step Solution:

1. **Understanding the Problem**:
The policeman detects a drop of 10% in the pitch of the horn as the car crosses him. This means the frequency of the sound decreases when the car moves away from him.

2. **Defining Frequencies**:
Let \( f \) be the original frequency of the horn. When the car approaches the policeman, the apparent frequency \( f_1 \) is given by:
\[ f_1 = f \cdot \frac{v}{v - v_s} \]
When the car moves away, the apparent frequency \( f_2 \) is:
\[ f_2 = f \cdot \frac{v}{v + v_s} \]
where:
- \( v \) = speed of sound = 330 m/s
- \( v_s \) = speed of the car (source)

3. **Setting Up the Relationship**:
According to the problem, \( f_2 \) is 90% of \( f_1 \):
\[ f_2 = 0.9 f_1 \]

4. **Substituting the Frequencies**:
Substitute the expressions for \( f_1 \) and \( f_2 \) into the equation:
\[ f \cdot \frac{v}{v + v_s} = 0.9 \left( f \cdot \frac{v}{v - v_s} \right) \]
Since \( f \) is common on both sides, we can cancel it out:
\[ \frac{v}{v + v_s} = 0.9 \cdot \frac{v}{v - v_s} \]

5. **Cross-Multiplying**:
Cross-multiply to eliminate the fractions:
\[ v(v - v_s) = 0.9 v(v + v_s) \]

6. **Expanding Both Sides**:
Expanding both sides gives:
\[ v^2 - v v_s = 0.9 v^2 + 0.9 v v_s \]

7. **Rearranging the Equation**:
Rearranging terms leads to:
\[ v^2 - 0.9 v^2 = v v_s + 0.9 v v_s \]
Simplifying further:
\[ 0.1 v^2 = 1.9 v_s v \]

8. **Solving for \( v_s \)**:
Divide both sides by \( v \) (assuming \( v \neq 0 \)):
\[ 0.1 v = 1.9 v_s \]
Thus:
\[ v_s = \frac{0.1 v}{1.9} \]

9. **Substituting the Value of \( v \)**:
Substitute \( v = 330 \, \text{m/s} \):
\[ v_s = \frac{0.1 \times 330}{1.9} = \frac{33}{1.9} \approx 17.37 \, \text{m/s} \]

10. **Final Answer**:
The speed of the car is approximately:
\[ v_s \approx 17.4 \, \text{m/s} \]