Amplitude of a mass-spring system, which is executing simple harmonic motion decreases with time. If mass $=500 \mathrm{~g}$, Decay constant $=20 \mathrm{~g} / \mathrm{s}$ then how much time is required for the amplitude of the system to drop to half of its initial value? ( $\ln 2=0.693$ )

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Elasticity & SHM

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 half of its initial value?
(

ln 2 = 0.693

)

Answer: (A)

In a damped mass-spring system the amplitude decays as

A = A_{0} e^{- b t /2 m}

with

b

the damping constant and

m

the mass. Setting

A = \frac{A _{0}}{2}

gives

e^{b t /2 m} = 2

, so

t = \frac{2 m ln 2}{b} = \frac{2 \times 0.5 \times 0.693}{0.02} = 34.65 s

using

m = 500 g = 0.5 kg

and

b = 20 g s^{- 1} = 0.02 kg s^{- 1}

. The amplitude falls to half its initial value in

34.65 s

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option