Question:
The displacement of a damped harmonic oscillator is given by
\(x(t)=e^{-0.1t}\cos(10\pi t+\phi)\).
Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to :
The displacement of a damped harmonic oscillator is given by
\(x(t)=e^{-0.1t}\cos(10\pi t+\phi)\).
Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to :
\(x(t)=e^{-0.1t}\cos(10\pi t+\phi)\).
Here t is in seconds. The time taken for its amplitude of vibration to drop to half of its initial value is close to :
Show Hint
The amplitude is \(e^{-0.1t}\); set it equal to 1/2 and solve for t.
Updated On: Nov 5, 2026
- 27 s
- 13 s
- 7 s
- 4 s
Show Solution
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The Correct Option is C
Solution and Explanation
The amplitude envelope is \(A(t)=e^{-0.1t}\) (initial amplitude 1).
Set \(e^{-0.1t}=\dfrac12\Rightarrow 0.1t=\ln2=0.693\Rightarrow t=6.93\ \mathrm{s}\approx7\) s.
Answer: (3) 7 s
Set \(e^{-0.1t}=\dfrac12\Rightarrow 0.1t=\ln2=0.693\Rightarrow t=6.93\ \mathrm{s}\approx7\) s.
Answer: (3) 7 s
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