Two sources \(A\) and \(B\) are producing notes of frequency \(680 \mathrm{~Hz}\). A listener moves from…
Two sources \(A\) and \(B\) are producing notes of frequency \(680 \mathrm{~Hz}\). A listener moves from \(A\) to \(B\) with a constant velocity \(v\). if the speed of sound in air is \(340 \mathrm{~ms}^{-1}\), the value of \(y\) so that he hears 10 beats per second is
\(2.0 \mathrm{~ms}^{-1}\)
\(2.5 \mathrm{~ms}^{-1}\)
\(3.0 \mathrm{~ms}^{-1}\)
\(3.5 \mathrm{~ms}^{-1}\)
Solution
Given,
notes of frequency produced by the sources \(A\) and \(B\) is \(680 \mathrm{~Hz}\).
i. e., \(f_A\) and \(f_B=680 \mathrm{~Hz}\)
Velocity of listener moves from \(A\) to \(B\) is constant \(=v\), speed of sound, \(v_s=340 \mathrm{~m} / \mathrm{s}\),
and beats per second, \(n=10\)
Now, beats per second from point \(A\) to \(B\) is given as
\(\begin{aligned}
n & =f_A\left(\frac{v_s+u}{v_s}\right)-f_B\left(\frac{v_s-u}{v_s}\right) \\
10 & =680\left(\frac{340+u}{340}\right)-680\left(\frac{340-u}{340}\right) \\
10 & =680\left[\left(\frac{340+u}{340}\right)-\left(\frac{340-u}{340}\right)\right] \\
1 & =68\left[\frac{340+u-340+u}{340}\right] \\
1 & =68\left(\frac{2 u}{340}\right) \\
\Rightarrow \quad 2 u & =\frac{340}{68}=\frac{340}{68 \times 2} \\
u & =2.5 \mathrm{~m} / \mathrm{s}
\end{aligned}\)