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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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Verified Answer
The correct answer is:
$3.0 \mathrm{~ms}^{-1}$
Listener go from $A \rightarrow B$ with velocity $(u)$ let the apparent frequency of sound from source $A$ by listener
or
$\begin{aligned}
n^{\prime} & =n\left(\frac{v-v_o}{v+v_s}\right) \\
n^{\prime} & =680\left(\frac{340-u}{340+0}\right)
\end{aligned}$
The apparent frequency of sound from source $B$ by listener
$\begin{aligned}
n^{\prime \prime} & =n\left(\frac{v+v_o}{v-v_s}\right) \\
& =680\left(\frac{340+u}{340-0}\right)
\end{aligned}$
But listener hear 10 beats per second.
Hence,
$n^{\prime \prime}-n^{\prime}=10$
or
$680\left(\frac{340+u}{340}\right)-680\left(\frac{340-u}{340}\right)=10$
or
$\begin{aligned}
\text{or} \quad 2(340+u-340+u) & =10 \\
\text{or} \quad u=2.5 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$
or
$\begin{aligned}
n^{\prime} & =n\left(\frac{v-v_o}{v+v_s}\right) \\
n^{\prime} & =680\left(\frac{340-u}{340+0}\right)
\end{aligned}$
The apparent frequency of sound from source $B$ by listener
$\begin{aligned}
n^{\prime \prime} & =n\left(\frac{v+v_o}{v-v_s}\right) \\
& =680\left(\frac{340+u}{340-0}\right)
\end{aligned}$
But listener hear 10 beats per second.
Hence,
$n^{\prime \prime}-n^{\prime}=10$
or
$680\left(\frac{340+u}{340}\right)-680\left(\frac{340-u}{340}\right)=10$
or
$\begin{aligned}
\text{or} \quad 2(340+u-340+u) & =10 \\
\text{or} \quad u=2.5 \mathrm{~m} \mathrm{~s}^{-1}
\end{aligned}$
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