Question

If the amplitudes of a damped harmonic oscillator at times 3 and 6 seconds are 6 cm and 4 cm respectively, then the initial amplitude of the oscillator is

• A. 20 cm
• B. 16 cm
• C. 9 cm
• D. 12 cm

Hint:

In damped harmonic motion, amplitude decays exponentially: \( A = A_0 e^{-bt} \). Use ratios of known amplitudes at two times to eliminate \( A_0 \) and solve for it.

Answer 1

The Correct Option is C

Step 1: Damped harmonic motion amplitude formula: \[ A(t) = A_0 e^{-bt} \] Given: \[ A(3) = 6 = A_0 e^{-3b}, \quad A(6) = 4 = A_0 e^{-6b} \] Divide the equations: \[ \frac{4}{6} = \frac{A_0 e^{-6b}}{A_0 e^{-3b}} = e^{-3b} \Rightarrow e^{-3b} = \frac{2}{3} \] Now substitute in: \[ 6 = A_0 \cdot \frac{2}{3} \Rightarrow A_0 = \frac{6 \cdot 3}{2} = 9\, \text{cm} \]