The amplitude of a damped oscillator decrease | vedclass

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

Question

(C) The amplitude of a damped oscillator is given by $A(t) = A_{0} e^{-\frac{bt}{2m}}$,where $A_{0}$ is the initial amplitude.According to the problem

Vedclass · Accepted Answer

$0.729$

Vedclass · Answer

(C) The amplitude of a damped oscillator is given by $A(t) = A_{0} e^{-\frac{bt}{2m}}$,where $A_{0}$ is the initial amplitude.According to the problem,after $t = 5\,s$,the amplitude becomes $0.9 A_{0}$:$0.9 A_{0} = A_{0} e^{-\frac{b(5)}{2m}}$$e^{-\frac{5b}{2m}} = 0.9$ ............... $(i)$We need to find the amplitude after another $10\,s$,which means at a total time $t = 5\,s + 10\,s = 15\,s$:$A(15) = A_{0} e^{-\frac{b(15)}{2m}}$$A(15) = A_{0} (e^{-\frac{5b}{2m}})^3$ ............... $(ii)$Substituting equation $(i)$ into equation $(ii)$:$A(15) = A_{0} (0.9)^3$$A(15) = A_{0} (0.729)$Thus,the amplitude decreases to $0.729$ times its original magnitude.Therefore,$\alpha = 0.729$.

The amplitude of a damped oscillator decreases to $0.9$ times its original magnitude in $5\,s$. In another $10\,s$ it will decrease to $\alpha$ times its original magnitude,where $\alpha$ equals

Similar Questions

The correct order of dipole moment is

The line $2x + \sqrt{6}y = 2$ is a tangent to the curve $x^2 - 2y^2 = 4$. The point of contact is

If a line $l$ passes through $(k, 2k), (3k, 3k)$ and $(3, 1)$ where $k \neq 0$,then the distance from the origin to the line $l$ is

If $|a| = 2$,$|b| = 3$ and $a, b$ are mutually perpendicular,then the area of the triangle whose vertices are $0, a + b, a - b$ is

The $emf$ (in $V$) of a Daniell cell containing $0.1 \ M$ $ZnSO_4$ and $0.01 \ M$ $CuSO_4$ solutions at their respective electrodes is
$(E_{Cu^{2+} / Cu}^{\circ} = +0.34 \ V; E_{Zn^{2+} / Zn}^{\circ} = -0.76 \ V)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module