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Q.

Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass=500g, Decay constant=20 g/s then how much time is required for the amplitude of the system to drop to half of its initial value?

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a

34.65s

b

0.034s.

c

15.01s.

d

17.32s

answer is B.

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Detailed Solution

$\begin{array}{l} A = A_{0} e^{- \frac{b t}{2 m}} \\ \Rightarrow \frac{A_{0}}{2} = A_{0} e^{- \frac{b t}{2 m}} \\ \Rightarrow \frac{b t}{2 m} = \ln 2 = 0.693 \\ \Rightarrow t = \frac{2 m}{b} \times 0.693 \\ \Rightarrow t = 2 \times \frac{500}{20} \times 0.693 = 50 \times 0.693 = 34.65 \sec . \end{array}$

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