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Distinction between interference and beats - class-XI
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Two sound sources are moving in opposite direction with velocity $v _1$ and $v _2$ $(v _1>v _2)$. Both are moving away from a stationary observer.the frequency of both the source is $900\ Hz$. What is the value of $v _1 - v _2 $ so that the beat frequency observed will be $6\ Hz$ ?
Speed of sound =$300\ ms^{-1}$
AnswerClick to flip back
A
$2\ ms^{-1}$
💡 Explanation:
$f _1 = 900(\dfrac{300}{300+v _1})$
$\implies f _1= 900(1+ \dfrac{v _1}{300})^{-1}$
$\implies f _1 = 900 - 3v _1$
Similarly
$f _2 = 900 - 3v _2$
So,
$f _1 -f _2 = 6$
$3(v _1 - v _2)= 6$
$\implies v _1 - v _2 = 2\ ms^{-1}$