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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Oscillations

An observer is moving towards a stationary source of frequency 320 Hz with a velocity of 10 m/s . The speed of sound in air is 343 m/s . The new frequency heard by the observer is

$(a)\;310 \;Hz\qquad(b)\;336 \;Hz\qquad(c)\;340 \;Hz\qquad(d)\;329 \;Hz$

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1 Answer

Answer : (d) $\;329 \;Hz$
Explanation :
New frequency $\;f^{|} =(\large\frac{v_{0}+v}{v})\;f$
$v_{0}\;$ velocity of observer
$v\;$ velocity of sound
By Doppler's effect
therefore , $\;f^{|}=(\large\frac{10+343}{343})\times320$
$=\large\frac{353}{343}\times320$
$=329\;Hz \;.$
answered Mar 8, 2014 by yamini.v
 

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