In damped oscillations, damping force is directly proportional to speed of oscillator. If amplitude becomes half of its maximum value in 1 s, then after 2 s amplitude will be (A0-. initial amplitude)

Question

In damped oscillations, damping force is directly proportional to speed of oscillator. If amplitude becomes half of its maximum value in 1 s, then after 2 s amplitude will be (A0-. initial amplitude) (A) (1/4) A0 (B) (1/2) A0 (C) A0 (D) (√3 A0/2)

Tardigrade Admin · Accepted Answer

A=A0 e-b t Amplitude becomes half hence (A0/2) A0 e-b t[t=1] ∴ e-b=(1/2) ∴ In two seconds A=A0((1/2))2 A=(A0/4)

Q. In damped oscillations, damping force is directly proportional to speed of oscillator. If amplitude becomes half of its maximum value in $1 s$ , then after $2 s$ amplitude will be $(A_{0} -$ initial amplitude)

$\frac{1}{4} A_{0}$

$\frac{1}{2} A_{0}$

$A_{0}$

$\frac{3 A _{0}}{2}$

$A = A_{0} e^{- b t}$
Amplitude becomes half hence
$\frac{A _{0}}{2} A_{0} e^{- b t} [t = 1]$
$∴ e^{- b} = \frac{1}{2}$
$∴$ In two seconds
$A = A_{0} (\frac{1}{2})^{2}$
$A = \frac{A _{0}}{4}$

Q. In damped oscillations, damping force is directly proportional to speed of oscillator. If amplitude becomes half of its maximum value in 1s, then after 2s amplitude will be (A0​− initial amplitude)

41​A0​

21​A0​

A0​

23​A0​​