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

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Aicharya · Accepted Answer

Listener go from $A ightarrow B$ with velocity $(u)$ let the apparent frequency of sound from source $A$ by listener or $\begin{aligned} n^{\prime} & =n\left(\frac{v-v_o}{v+v_s}ight) \ n^{\prime} & =680\left(\frac{340-u}{340+0}ight) \end{aligned}$ The apparent frequency of sound from source $B$ by listener $\begin{aligned} n^{\prime \prime} & =n\left(\frac{v+v_o}{v-v_s}ight) \ & =680\left(\frac{340+u}{340-0}ight) \end{aligned}$ But listener hear 10 beats per second. Hence, $n^{\prime \prime}-n^{\prime}=10$ or $680\left(\frac{340+u}{340}ight)-680\left(\frac{340-u}{340}ight)=10$ or $\begin{aligned} 	ext{or} \quad 2(340+u-340+u) & =10 \ 	ext{or} \quad u=2.5 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned}$

Two sources $A$ and $B$ are sending notes of frequency $680 \mathrm{~Hz}$. A listener moves from $A$ and $B$…

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