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JEE Advanced 2025Paper 2

Exam: JEE Advanced
Year: 2025
Paper: Paper 2
Subject: Physics
Chapter: Waves and Sound

JEE Advanced 2025 · Paper 2 · Physics · Waves and Sound

An audio transmitter (T) and a receiver (R) are hung vertically...

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

An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings of length 8 m with

Correct answer

31.19

Step-by-step solution

\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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