Question:
The amplitude of a damped harmonic oscillator becomes \( \frac{1}{n} \) times its initial amplitude \( A_0 \) at the end of 20 oscillations. The amplitude of the oscillator when it completes 40 oscillations is:
The amplitude of a damped harmonic oscillator becomes \( \frac{1}{n} \) times its initial amplitude \( A_0 \) at the end of 20 oscillations. The amplitude of the oscillator when it completes 40 oscillations is:
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The damping factor in an oscillatory system affects the amplitude, and it decreases exponentially over time.
Updated On: May 5, 2026
- \( \frac{A_0}{n^3} \)
- \( A_0 \)
- \( \frac{A_0}{n^2} \)
- \( \frac{A_0}{n} \)
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The Correct Option is C
Solution and Explanation
- In a damped oscillator, amplitude varies with time as: \[ A = A_0 e^{-kt} \]
- Time for one oscillation is constant, so after \(N\) oscillations: \[ A = A_0 e^{-kN T} \]
- Given after 20 oscillations: \[ \frac{A}{A_0} = e^{-20kT} = \frac{1}{n} \]
- Taking square on both sides: \[ e^{-40kT} = \left(\frac{1}{n}\right)^2 \]
- After 40 oscillations: \[ A = A_0 e^{-40kT} = \frac{A_0}{n^2} \]
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