Two sources A and B are sending notes of freq | vedclass

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(B) Let the frequency of the sources be $n = 680 \ Hz$ and the speed of sound be $v = 340 \ ms^{-1}$.The listener moves from $A$ towards $B$ with velo

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(B) Let the frequency of the sources be $n = 680 \ Hz$ and the speed of sound be $v = 340 \ ms^{-1}$.The listener moves from $A$ towards $B$ with velocity $u$.The apparent frequency $n'$ heard by the listener from source $A$ (moving away from $A$) is:$n' = n \left( \frac{v - u}{v} \right) = 680 \left( \frac{340 - u}{340} \right) = 2(340 - u)$The apparent frequency $n''$ heard by the listener from source $B$ (moving towards $B$) is:$n'' = n \left( \frac{v + u}{v} \right) = 680 \left( \frac{340 + u}{340} \right) = 2(340 + u)$The beat frequency is the difference between the two apparent frequencies:$|n'' - n'| = 10$$2(340 + u) - 2(340 - u) = 10$$680 + 2u - 680 + 2u = 10$$4u = 10$$u = 2.5 \ ms^{-1}$

Two sources $A$ and $B$ are sending notes of frequency $680 \ Hz$. $A$ listener moves from $A$ towards $B$ with a constant velocity $u$. If the speed of sound in air is $340 \ ms^{-1}$, what must be the value of $u$ so that he hears $10$ beats per second (in $ms^{-1}$)?

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