The amplitude of a damped oscillator becomes | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The amplitude of a damped oscillator at time $t$ is given by $A(t) = A_0 e^{-\frac{bt}{2m}}$.Given that at $t = 2 \, s$,the amplitude becomes $1/3

Vedclass · Accepted Answer

$3^3$

Vedclass · Answer

(D) The amplitude of a damped oscillator at time $t$ is given by $A(t) = A_0 e^{-\frac{bt}{2m}}$.Given that at $t = 2 \, s$,the amplitude becomes $1/3$ of the original amplitude $A_0$:$\frac{A_0}{3} = A_0 e^{-\frac{b(2)}{2m}}$$\frac{1}{3} = e^{-\frac{b}{m}}$Now,we need to find the amplitude at $t = 6 \, s$:$A(6) = A_0 e^{-\frac{b(6)}{2m}} = A_0 (e^{-\frac{b}{m}})^3$Substituting the value $e^{-\frac{b}{m}} = 1/3$:$A(6) = A_0 \left(\frac{1}{3}\right)^3 = \frac{A_0}{27}$Given that $A(6) = \frac{A_0}{n}$,we have:$\frac{A_0}{n} = \frac{A_0}{27}$$n = 27 = 3^3$.

The amplitude of a damped oscillator becomes one third in $2 \, s$. If its amplitude after $6 \, s$ is $1/n$ times the original amplitude,then the value of $n$ is

Similar Questions

In damped $SHM$,the $SI$ unit of damping constant is

$A$ block of mass $200 \, g$ is executing $SHM$ under the influence of a spring with spring constant $K = 90 \, N \, m^{-1}$ and a damping constant $b = 40 \, g \, s^{-1}$. The time elapsed for its amplitude to drop to half of its initial value is ...... $s$ (Given $\ln \frac{1}{2} = -0.693$).

Which of the following figures represents damped harmonic motion?

In forced oscillations,a particle oscillates simple harmonically with a frequency equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module