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A car is moving with 90 km h^(-1) blows ...

A car is moving with `90 km h^(-1)` blows a horn of 150 Hz, towards a cliff. The frequency of the reflected sound heard by the driver will be (speed of sound in air is `340 ms^(-1)` )

A

150 Hz

B

140 Hz

C

180 Hz

D

174 Hz

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the frequency of the reflected sound heard by the driver of the car moving towards a cliff. We'll use the Doppler effect formula for sound. ### Step-by-Step Solution: 1. **Convert the speed of the car from km/h to m/s:** \[ \text{Speed of the car} = 90 \text{ km/h} = 90 \times \frac{1000 \text{ m}}{3600 \text{ s}} = 25 \text{ m/s} \] **Hint:** Remember that to convert km/h to m/s, multiply by \( \frac{5}{18} \). 2. **Identify the parameters:** - Speed of sound in air, \( v = 340 \text{ m/s} \) - Frequency of the horn, \( f = 150 \text{ Hz} \) - Speed of the observer (the driver), \( v_o = 25 \text{ m/s} \) (since the car is moving towards the cliff) - Speed of the source (the cliff), \( v_s = 0 \text{ m/s} \) (the cliff is stationary) 3. **Calculate the apparent frequency when the sound reaches the cliff:** Using the Doppler effect formula: \[ f' = f \frac{v + v_o}{v - v_s} \] Substituting the values: \[ f' = 150 \frac{340 + 25}{340 - 0} = 150 \frac{365}{340} \] **Hint:** When calculating the apparent frequency, ensure you correctly identify which object is the source and which is the observer. 4. **Calculate \( f' \):** \[ f' = 150 \times \frac{365}{340} = 150 \times 1.0735 \approx 160.98 \text{ Hz} \] 5. **Now, this frequency \( f' \) acts as the source frequency when the sound reflects back to the car. Calculate the frequency heard by the driver:** Using the Doppler effect formula again, but now the car is the observer and the cliff is the source: \[ f'' = f' \frac{v + v_o}{v - v_s} \] Substituting the values: \[ f'' = 160.98 \frac{340 + 25}{340 - 0} = 160.98 \frac{365}{340} \] **Hint:** Remember that the roles of source and observer switch when the sound reflects. 6. **Calculate \( f'' \):** \[ f'' = 160.98 \times \frac{365}{340} \approx 160.98 \times 1.0735 \approx 172.87 \text{ Hz} \] 7. **Final Result:** Rounding to the nearest whole number, the frequency of the reflected sound heard by the driver is approximately: \[ f'' \approx 173 \text{ Hz} \] ### Conclusion: The frequency of the reflected sound heard by the driver will be approximately **173 Hz**.
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