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JEE Advanced · Physics · 16. Waves & Sound

An audio transmitter \((T)\) and a receiver \((R)\) are hung vertically from two identical massless strings of length 8 m with their pivots well separated along the \(X\) axis. They are pulled from the equilibrium position in opposite directions along the \(X\) axis by a small angular amplitude \(\theta_0=\cos ^{-1}(0.9)\) and released simultaneously. If the natural frequency of the transmitter is 660 Hz and the speed of sound in air is \(330 \mathrm{~m} / \mathrm{s}\), the maximum variation in the frequency (in Hz ) as measured by the receiver (Take the acceleration due to gravity \(g=10 \mathrm{~m} / \mathrm{s}^2\) ) is \(\qquad\)

  1. A 32
  2. B 34
  3. C 33
  4. D 35
Verified Solution

Answer & Solution

Correct Answer

(A) 32

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & \cos \theta_0=1-\frac{\theta_0^2}{2}=0.9 \\ & \frac{\theta_0^2}{2}=0.1 \Rightarrow \theta_0=10.2=\frac{1}{\sqrt{5}}\end{aligned}\)

\(\begin{aligned} & f_{\max }=\frac{v+v^{\prime}}{v-v^{\prime}} f \\ & f_{\min }=\frac{v-v^{\prime}}{v+v^{\prime}} f \\ & \Delta f_{\max }=f_{\max }-f_{\min }=\frac{v+v^{\prime}}{v-v^{\prime}} f-\frac{v-v^{\prime}}{v+v^{\prime}} f \\ & =\frac{\left(v+v^{\prime}\right)^2-\left(v-v^{\prime}\right)^2}{v^2-v^{\prime 2}} f\end{aligned}\)
\(\Delta f_{\max }=\frac{4 v v^{\prime}}{v^2-v^{\prime 2}} f \qquad ....(i)\)
Here, \(\mathrm{v}^{\prime}=\ell \Omega_{\max }\)
\(=\ell \cdot \theta_0 \cdot \omega \quad(\omega=\text { angular frequency })\)
\(\begin{aligned} & =\ell \theta_0 \sqrt{\frac{\mathrm{~g}}{\ell}} \\ & \mathrm{v}^{\prime}=\theta_0 \sqrt{\mathrm{~g} \ell} \\ & \mathrm{v}^{\prime}=\frac{1}{\sqrt{5}} \sqrt{10 \times 8} \\ & \mathrm{v}^{\prime}=4\end{aligned}\)
Put in equation (i)
\(\begin{aligned} & \Delta \mathrm{f}_{\max }=\frac{4 \times 330 \times 4 \times 660}{330^2-4^2} \\ & \approx \frac{16 \times 330 \times 660}{330} \approx 32\end{aligned}\)
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