The displacement of a damped harmonic oscillator is given by \[ x(t)=e^{-0.1 t} \cos (10 \pi t+\varphi) \text {. Here } t \text { is in seconds. }\] The time taken for its amplitude of vibration to drop to half of its initial value is close to:

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Elasticity & SHM

The displacement of a damped harmonic oscillator is given by

x (t) = e^{- 0.1 t} cos (10 π t + φ) . Here t is in seconds.

The time taken for its amplitude of vibration to drop to half of its initial value is close to:

Answer: (D)

The amplitude of the damped oscillator decays as

A (t) = A_{0} e^{- 0.1 t}

. Setting it to half the initial value,

e^{- 0.1 t} = \frac{1}{2}

, so

- 0.1 t = - ln 2

and

t = \frac{l n 2}{0.1} = 10 ln 2 \approx 6.93

s, close to

7

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option