Question

The amplitude of a damped harmonic oscillator becomes 50% of its initial value in a time of 12 s. If the amplitude of the oscillator at a time of 36 s is \(x%\) of its initial amplitude, then the value of \(x\) is

• A. \(25\)
• B. \(12.5\)
• C. \(37.5\)
• D. \(8\)

Hint:

In damped harmonic motion, the amplitude decreases exponentially with time. If it becomes half in a fixed interval, after thrice that time, it becomes \((1/2)^3 = 1/8\) of the original.

Answer 1

The Correct Option is B

Step 1: Damped amplitude decreases exponentially: \[ A = A_0 e^{-bt} \] Step 2: Given: at \(t = 12\,\text{s}\), amplitude becomes \(50%\): \[ 0.5 A_0 = A_0 e^{-12b} \Rightarrow e^{-12b} = 0.5 \] Step 3: At \(t = 36\,\text{s}\), \[ A = A_0 e^{-36b} = A_0 \left(e^{-12b}\right)^3 = A_0 (0.5)^3 = A_0 \times 0.125 \] Step 4: Percentage of initial amplitude: \[ x = 0.125 \times 100 = 12.5 \] % Final Answer \[ \boxed{12.5} \]