Question

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 m/s, the maximum variation in the frequency (in Hz) as measured by the receiver (Take the acceleration due to gravity $g = 10\, \text{m/s}^2$) is ___

Hint:

Maximum velocity in pendulum oscillation is \(v_{\max} = \omega_0 L \sin \theta_0\). Doppler frequency shift depends on relative velocity over speed of sound. For opposite motions, relative velocity doubles.

Answer 1

Correct Answer: 31

Step 1: Determine the angular frequency of oscillation The angular frequency of a simple pendulum is given by $\omega = \sqrt{\frac{g}{L}}$. Given $g = 10 \, \text{m/s}^2$ and $L = 8 \, \text{m}$, $$ \omega = \sqrt{\frac{10}{8}} = \sqrt{1.25} = \frac{\sqrt{5}}{2} \, \text{rad/s} $$
Step 2: Determine the maximum linear velocity of the transmitter and receiver The maximum angular displacement is $\theta_0 = \cos^{-1}(0.9)$. $$ \sin^2 \theta_0 = 1 - \cos^2 \theta_0 = 1 - (0.9)^2 = 1 - 0.81 = 0.19 $$ $$ \sin \theta_0 = \sqrt{0.19} $$ The maximum linear velocity $v_{max}$ in SHM is approximately $v_{max} = L \omega_{max}$, where $\omega_{max} \approx \theta_0 \omega$. However, considering the horizontal velocity component, the maximum horizontal velocity is $v_{max} \approx L \omega \sin \theta_0$. $$ v_{max} = 8 \times \frac{\sqrt{5}}{2} \times \sqrt{0.19} = 4 \sqrt{0.95} \, \text{m/s} $$ Approximating $\sqrt{0.95} \approx 0.97468$: $$ v_{max} \approx 4 \times 0.97468 \approx 3.8987 \, \text{m/s} $$
Step 3: Apply the Doppler effect formula The Doppler effect formula is $f' = f \left( \frac{v_{sound} \pm v_{receiver}}{v_{sound} \mp v_{source}} \right)$. Given $f = 660 \, \text{Hz}$ and $v_{sound} = 330 \, \text{m/s}$. For maximum observed frequency ($f'_{max}$), the receiver moves towards the source and the source moves away from the receiver: $$ f'_{max} = 660 \left( \frac{330 + v_{max}}{330 - v_{max}} \right) = 660 \left( \frac{330 + 3.8987}{330 - 3.8987} \right) \approx 660 \left( \frac{333.8987}{326.1013} \right) \approx 675.47 \, \text{Hz} $$ For minimum observed frequency ($f'_{min}$), the receiver moves away from the source and the source moves towards the receiver: $$ f'_{min} = 660 \left( \frac{330 - v_{max}}{330 + v_{max}} \right) = 660 \left( \frac{330 - 3.8987}{330 + 3.8987} \right) \approx 660 \left( \frac{326.1013}{333.8987} \right) \approx 644.53 \, \text{Hz} $$ The maximum variation in frequency is: $$ \Delta f = f'_{max} - f'_{min} = 675.47 - 644.53 = 30.94 \, \text{Hz} $$ Rounding to the nearest integer, the maximum variation is $31 \, \text{Hz}$.