Question

The amplitude of a damped harmonic oscillator becomes \[ \frac{1}{2\sqrt2} \] times its initial amplitude in a time of \(6\) s. Additional time taken for the amplitude to become \[ \frac{1}{4\sqrt2} \] times its initial amplitude is

• A. \(4\) s
• B. \(6\) s
• C. \(8\) s
• D. \(10\) s

Hint:

In damped oscillations, \[ \boxed{A=A_0e^{-kt}.} \] Take the ratio of amplitudes at different times to eliminate the initial amplitude and determine the required time interval.

Answer 1

The Correct Option is A

Step 1: Use the equation for amplitude of damped oscillation. The amplitude varies as \[ A=A_0e^{-kt}, \] where \(k\) is the damping constant. Given, \[ A=A_0\left(\frac1{2\sqrt2}\right) \] at \[ t=6\text{ s}. \] Hence, \[ e^{-6k} = \frac1{2\sqrt2} = \frac1{2^{3/2}}. \] Taking logarithms, \[ 6k=\frac32\ln2, \] so \[ k=\frac14\ln2. \]

Step 2: Find the total time for the new amplitude. Now, \[ A=A_0\left(\frac1{4\sqrt2}\right) = \frac1{2^{5/2}}A_0. \] Thus, \[ e^{-kt} = \frac1{2^{5/2}}. \] Hence, \[ kt=\frac52\ln2, \] which gives \[ t = \frac{\frac52\ln2}{\frac14\ln2} = 10\text{ s}. \]

Step 3: Find the additional time. Additional time required is \[ 10-6=4\text{ s}. \] Therefore, \[ \boxed{4\text{ s}}. \] Thus, \[ \boxed{(A)} \] is the correct answer.