Apparent Frequency Heard by Car Driver Moving Towards Building

To find the frequency of the sound (reflected from the building) heard by the car driver, we can apply the principles of the Doppler Effect in two steps:

1. Frequency Received by the Building

The building acts as a stationary observer ($v_{\text{building}} = 0$) receiving sound from the car, which is a source moving towards it with velocity $v_s$.

Given:

• Original frequency of the siren, $f_0 = 8\text{ kHz}$
• Velocity of the car, $v_s = 36\text{ km/hr} = 36 \times \frac{5}{18}\text{ m/s} = 10\text{ m/s}$
• Speed of sound in air, $v = 320\text{ m/s}$

The frequency $f'$ of the sound reaching the building is given by:
$f' = f_0 \left( \frac{v}{v - v_s} \right)$

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2. Frequency Heard by the Driver (Reflected Sound)

The building reflects this sound at frequency $f'$, acting as a stationary source. The car driver is an observer moving towards this stationary source with speed $v_o = v_s = 10\text{ m/s}$.

The frequency $f''$ heard by the driver is:
$f'' = f' \left( \frac{v + v_o}{v} \right)$

Substituting the expression for $f'$ from Step 1:
$f'' = f_0 \left( \frac{v}{v - v_s} \right) \left( \frac{v + v_s}{v} \right) = f_0 \left( \frac{v + v_s}{v - v_s} \right)$

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3. Calculation

Substitute the given values into the equation:
$f'' = 8\text{ kHz} \times \left( \frac{320 + 10}{320 - 10} \right)$

$f'' = 8 \times \left( \frac{330}{310} \right) = 8 \times \left( \frac{33}{31} \right)\text{ kHz}$

$f'' = \frac{264}{31}\text{ kHz} \approx 8.516\text{ kHz} \approx 8.50\text{ kHz}$

Thus, the frequency of the siren heard by the car driver is approximately $8.50\text{ kHz}$.

Correct Option: A