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Q. The source of sound generating a frequency of $3 \,kHz$ reaches an observer with a speed of $0.5$ times the velocity of sound in air. The frequency heard by the observer is

BITSATBITSAT 2011

Solution:

Apparent frequency heard will be $n '= n \frac{n}{v-v_{s}}$
$v$ = velocity of sound,
$v_{s}=$ velocity of source of sound
$n =$ frequency $-3 KHz$
$\therefore n'=3 \times \frac{v}{v-0.5_{v}}$
$=3 \times \frac{v}{0.5_{v}}=6 \,k H z$