# 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

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

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 , $\;f^{|}=(\large\frac{10+343}{343})\times320$
$=\large\frac{353}{343}\times320$
$=329\;Hz \;.$