A motorcyclist is approaching a large wall wi | vedclass

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(C) First,convert velocities to $m/s$:Motorcyclist velocity $v_m = 90 	imes \frac{5}{18} = 25\,m/s$.Car velocity $v_c = 108 	imes \frac{5}{18} = 30\

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(C) First,convert velocities to $m/s$:Motorcyclist velocity $v_m = 90 	imes \frac{5}{18} = 25\,m/s$.Car velocity $v_c = 108 	imes \frac{5}{18} = 30\,m/s$.Speed of sound $v = 330\,m/s$.Frequency of horn $f_s = 30\,Hz$.$1$. Frequency heard directly from the car $(f_1)$:The car is the source $(v_s = 30\,m/s)$ and the motorcyclist is the observer $(v_o = 25\,m/s)$. Both are moving in the same direction towards the wall.$f_1 = f_s \left( \frac{v - v_o}{v - v_s} ight) = 30 \left( \frac{330 - 25}{330 - 30} ight) = 30 \left( \frac{305}{300} ight) = 30.5\,Hz$.$2$. Frequency heard after reflection from the wall $(f_2)$:The wall acts as a stationary source reflecting the sound. The sound first travels to the wall at $f' = f_s \left( \frac{v}{v - v_s} ight)$. Then the wall reflects this to the observer moving towards the wall at $v_o = 25\,m/s$.$f_2 = f' \left( \frac{v + v_o}{v} ight) = f_s \left( \frac{v}{v - v_s} ight) \left( \frac{v + v_o}{v} ight) = f_s \left( \frac{v + v_o}{v - v_s} ight) = 30 \left( \frac{330 + 25}{330 - 30} ight) = 30 \left( \frac{355}{300} ight) = 35.5\,Hz$.$3$. Beat frequency:$f_{	ext{beat}} = |f_2 - f_1| = 35.5 - 30.5 = 5\,Hz$.

$A$ motorcyclist is approaching a large wall with a velocity of $90\,km/hr$. $A$ car is chasing him with a velocity of $108\,km/hr$. If the car sounds a horn at $30\,Hz,$ the beat frequency heard by the motorcyclist is . . . . . . $Hz$. Take the velocity of sound $= 330\,m/s$.

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