At what speed should a car approach a stationary observer if the hears the music of the car radio with a frequency 20% higher than it actully is ? The speed of sound is `340 ms^(-1)`

Answer

To solve the problem, we need to find the speed at which a car should approach a stationary observer so that the frequency of the music heard by the observer is 20% higher than the actual frequency of the music played in the car.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Speed of sound, \( v = 340 \, \text{m/s} \)
- Frequency increase: The observed frequency \( f_2 \) is 20% higher than the actual frequency \( f_1 \).
- Therefore, if we assume the actual frequency \( f_1 = 100 \, \text{Hz} \), then:
\[ f_2 = f_1 + 0.2 f_1 = 1.2 f_1 = 120 \, \text{Hz} \]

2. **Use the Doppler Effect Formula:**
- The formula for the Doppler effect when the source is moving towards a stationary observer is:
\[ f_2 = f_1 \frac{v}{v - v_s} \]
- Here, \( v_s \) is the speed of the source (the car) and \( v_0 \) (the speed of the observer) is 0 since the observer is stationary.

3. **Substituting the Known Values:**
- Substitute \( f_2 = 120 \, \text{Hz} \), \( f_1 = 100 \, \text{Hz} \), and \( v = 340 \, \text{m/s} \) into the formula:
\[ 120 = 100 \frac{340}{340 - v_s} \]

4. **Rearranging the Equation:**
- Multiply both sides by \( (340 - v_s) \):
\[ 120 (340 - v_s) = 100 \cdot 340 \]
- Simplifying gives:
\[ 40800 - 120 v_s = 34000 \]

5. **Solving for \( v_s \):**
- Rearranging the equation:
\[ 40800 - 34000 = 120 v_s \]
\[ 6800 = 120 v_s \]
- Dividing both sides by 120:
\[ v_s = \frac{6800}{120} = 56.67 \, \text{m/s} \]

### Final Answer:
The speed at which the car should approach the stationary observer is approximately \( 56.67 \, \text{m/s} \).

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