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(C) The engine acts as both the source and the observer. The sound travels to the wall and reflects back.Since the engine is moving towards the wall w

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$1.6$

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(C) The engine acts as both the source and the observer. The sound travels to the wall and reflects back.Since the engine is moving towards the wall with velocity $v_s = 50\, ms^{-1}$,the frequency $f'$ received by the wall is given by the Doppler effect formula for a moving source and stationary observer:$f' = f_0 \left( \frac{v}{v - v_s} ight) = 1.2 \left( \frac{350}{350 - 50} ight) = 1.2 \left( \frac{350}{300} ight) = 1.4\, kHz$.Now,the wall acts as a stationary source reflecting this frequency $f'$,and the engine acts as an observer moving towards the wall with velocity $v_o = 50\, ms^{-1}$.The frequency $f''$ heard by the driver is:$f'' = f' \left( \frac{v + v_o}{v} ight) = 1.4 \left( \frac{350 + 50}{350} ight) = 1.4 \left( \frac{400}{350} ight) = 1.4 	imes \frac{8}{7} = 1.6\, kHz$.Alternatively,using the combined formula for reflection from a wall:$f'' = f_0 \left( \frac{v + v_s}{v - v_s} ight) = 1.2 \left( \frac{350 + 50}{350 - 50} ight) = 1.2 \left( \frac{400}{300} ight) = 1.6\, kHz$.

An engine is moving towards a wall with a velocity $50\, ms^{-1}$ and emits a note of $1.2\, kHz$. The speed of sound in air is $350\, ms^{-1}$. The frequency of the note after reflection from the wall as heard by the driver of the engine is ..... $kHz$.

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