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 ...... \(s\) time is required for the amplitude of the system to drop to half of its initial value ? \((\ln 2=0.693)\)

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Answer & Solution

Correct Answer

(A) \(34.65\)

Step-by-step Solution

Detailed explanation

\(A = A _{0} e ^{-\gamma t }= A _{0} e ^{-\frac{ bt }{2 m }}\)

\(\frac{ A _{0}}{2}= A _{0} e ^{-\frac{ bt }{2 m }}\)

\(\frac{ bt }{2 m }=\ln 2\)

\(t =\frac{2 m }{ b } \ln 2=\frac{2 \times 500 \times 0.693}{20}\)

\(t =34.65\, second.\)

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