The displacement of a damped harmonic oscillator is given by x(t ) = e-01.1t 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 :

Question

The displacement of a damped harmonic oscillator is given by x(t ) = e-01.1t 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 : (A) 13 s (B) 7 s (C) 27 s (D) 4 s

Tardigrade Admin · Accepted Answer

A = A0 e-0.1t = (A0/2) ln 2 = 0.1 t t = 10 ln2 = 6.93 ≈ 7 sec

Q. The displacement of a damped harmonic oscillator is given by $x (t) = e^{- 01.1 t} cos (10 π t + ϕ)$ . Here $t$ is in seconds.
The time taken for its amplitude of vibration to drop to half of its initial value is close to :

13 s

7 s

27 s

4 s

$A = A_{0} e^{- 0.1 t} = \frac{A _{0}}{2}$
$l n 2 = 0.1 t$
$t = 10 l n 2 = 6.93 \approx 7 sec$

Q. The displacement of a damped harmonic oscillator is given by x(t)=e−01.1tcos(10πt+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 :

13 s

7 s

27 s

4 s

A=A0​e−0.1t=2A0​​ ln2=0.1t t=10ln2=6.93≈7sec

Q. The displacement of a damped harmonic oscillator is given by $x (t) = e^{- 01.1 t} cos (10 π t + ϕ)$ . Here $t$ is in seconds.
The time taken for its amplitude of vibration to drop to half of its initial value is close to :

$A = A_{0} e^{- 0.1 t} = \frac{A _{0}}{2}$
$l n 2 = 0.1 t$
$t = 10 l n 2 = 6.93 \approx 7 sec$