A soundof frequency 512 Hz is emitted from a stationary source . A train running at a speed of 72 km . h^(-1)passes the source . What will be the frequency of the sound heard by a passenger of the train before and after passing the source ? Neglect the effect of wind . Velocity of sound is 336 .S^(-1) .

A soundof frequency 512 Hz is emitted from a stationary source . A tra

1.	A soundof frequency 512 Hz is emitted from a stationary source . A train running at a speed of 72 km . h^(-1)passes the source . What will be the frequency of the sound heard by a passenger of the train before and after passing the source ? Neglect the effect of wind . Velocity of sound is 336 .S^(-1) .
Answer» Solution :Velocity of the train = 72 km . ` h^(-1)` ` = (72 xx1000)/(60xx60) m . S^(-1) = 20 m. S^(-1)` When the train approaches the source , the velocity of sound RELATIVE to the passenger , ` V. = 336 + 20 = 356 m . S^(-1)` SINCE the source is at REST , the wavelength of sound remains the same . ` THEREFORE` Apparent freuency due to Doppler effect `(V.) /(lambda ) = (V.) /((V)/(n)) = (V. n)/(V) = (356 xx 512)/(336) = 542 .5 Hz ` Again when the train recedes from the source , the velocity of sound relative to the passenger ` Again when the train recedes from the source , the velocity of sound relative to the passenger ` V. =336 - 20 = 316 . s^(-1) ` ` thereforeApparent freaquency ` = (V. n)/(V) = (316 xx 512)/(336) = 481 .5 Hz `