Question 16

NUMERICALHARD

An audio transmitter (T) and a receiver (R) are hung vertically from two identical massless strings of length 88 m with their pivots well separated along the XX axis. They are pulled from the equilibrium position in opposite directions along the XX axis by a small angular amplitude θ0=cos⁡−1(0.9)\theta_0 = \cos^{-1}(0.9) and released simultaneously. If the natural frequency of the transmitter is 660660 Hz and the speed of sound in air is 330330 m/s, the maximum variation in the frequency (in Hz) as measured by the receiver (Take the acceleration due to gravity g=10g = 10 m/s2^2) is ___

Question
Correct Answer: 32

Detailed Solution

  1. Maximum speed of pendulum: From conservation of energy, vmax=2gL(1−cos⁡θ0)=2×10×8×(1−0.9)=16=4v_{max} = \sqrt{2gL(1 - \cos\theta_0)} = \sqrt{2 \times 10 \times 8 \times (1 - 0.9)} = \sqrt{16} = 4 m/s.
  2. Both TT and RR have vmax=4v_{max} = 4 m/s at their respective equilibrium positions. Since they are released simultaneously in opposite directions, they move towards each other with maximum speed and away from each other with maximum speed.
  3. Maximum observed frequency (approaching): fmax=f0v+vRv−vT=660330+4330−4=660×334326≈676.2f_{max} = f_0 \frac{v + v_R}{v - v_T} = 660 \frac{330 + 4}{330 - 4} = 660 \times \frac{334}{326} \approx 676.2 Hz.
  4. Minimum observed frequency (receding): fmin=f0v−vRv+vT=660330−4330+4=660×326334≈644.2f_{min} = f_0 \frac{v - v_R}{v + v_T} = 660 \frac{330 - 4}{330 + 4} = 660 \times \frac{326}{334} \approx 644.2 Hz.
  5. Maximum variation: Δf=fmax−fmin≈676.2−644.2=32\Delta f = f_{max} - f_{min} \approx 676.2 - 644.2 = 32 Hz.
Free Exam

Boost Your Exam Preparation!

Move beyond just reading solutions. Access our comprehensive Test Series, original Mock Tests, and interactive learning modules. Many premium tests are completely free!

  • Original Mocks & Regular Test Series
  • Real NTA-like Interface with Analytics
  • Many Free Tests Available