The displacement of a damped harmonic oscillator is given by x(t)=e- 0 . 1 tcos(10 π t + varphi) , where t is the time in seconds. The time taken for the amplitude of the oscillator 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- 0 . 1 tcos(10 π t + varphi) , where t is the time in seconds. The time taken for the amplitude of the oscillator to drop to half of its initial value is close to (A) 13s (B) 27s (C) 4 s (D) 7s

Tardigrade Admin · Accepted Answer

(A0/2)=A0e- 0 . 1 t⇒ e0 . 1 t=2⇒ 0.1t=ln2 Therefore, t=(ln 2/0 . 1)=10ln2=6.93≈7s

Q. The displacement of a damped harmonic oscillator is given by $x (t) = e^{- 0.1 t} cos (10 π t + φ)$ , where $t$ is the time in seconds. The time taken for the amplitude of the oscillator to drop to half of its initial value is close to

$13 s$

$27 s$

$4 s$

$7 s$

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

Q. The displacement of a damped harmonic oscillator is given by x(t)=e−0.1tcos(10πt+φ) , where t is the time in seconds. The time taken for the amplitude of the oscillator to drop to half of its initial value is close to

13s

27s

4s

7s