Question:
Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500 g, Decay constant = 20 g/s then how much time is required for the amplitude of the system to drop to drop to half of its initial value? (ln 2 = 0.693)
Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass = 500 g, Decay constant = 20 g/s then how much time is required for the amplitude of the system to drop to drop to half of its initial value? (ln 2 = 0.693)
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Damped SHM amplitude is \(A = A_0 e^{-bt/2m}\); put \(A = A_0/2\) and solve for t.
Updated On: Nov 4, 2026
- 17.32 s
- 34.65 s
- 0.034 s
- 15.01 s
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The Correct Option is B
Solution and Explanation
For damped oscillations the amplitude decays as \(A = A_0 e^{-bt/(2m)}\).
Given \(m = 500\) g, \(b = 20\) g/s, \(A = A_0/2\).
\(\frac{1}{2} = e^{-bt/(2m)} \Rightarrow \ln 2 = \frac{bt}{2m}\).
\(t = \frac{2m\ln 2}{b} = \frac{2 \times 500 \times 0.693}{20} = 34.65\) s.
Answer: (2) 34.65 s
Given \(m = 500\) g, \(b = 20\) g/s, \(A = A_0/2\).
\(\frac{1}{2} = e^{-bt/(2m)} \Rightarrow \ln 2 = \frac{bt}{2m}\).
\(t = \frac{2m\ln 2}{b} = \frac{2 \times 500 \times 0.693}{20} = 34.65\) s.
Answer: (2) 34.65 s
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