Question

If the amplitude of a damped harmonic oscillator becomes half of its initial amplitude in a time of 10 s, then the time taken for the mechanical energy of the oscillator to become half of its initial mechanical energy is

• A. 2.5 s
• B. 20 s
• C. 10 s
• D. 5 s

Hint:

The energy of a damped oscillator decays exponentially twice as fast as the amplitude. This is because \(E \propto A^2\) and \(A \propto e^{-\gamma t}\), so \(E \propto (e^{-\gamma t})^2 = e^{-2\gamma t}\). This means the time constant for energy decay is half the time constant for amplitude decay. So, if amplitude halves in time T, energy halves in time T/2. Here, T=10s, so the answer is 10/2 = 5s.

Answer 1

The Correct Option is D

Step 1: Amplitude Decay Formula: For a damped harmonic oscillator, amplitude decays exponentially: \[ A(t) = A_0 e^{-\gamma t} \] (where \( \gamma = b/2m \) is the decay constant for amplitude). Given that at \( t = 10 \, \text{s} \), \( A = A_0/2 \). \[ \frac{A_0}{2} = A_0 e^{-10\gamma} \implies e^{-10\gamma} = \frac{1}{2} \]
Step 2: Energy Decay Formula: The mechanical energy is proportional to the square of the amplitude (\( E \propto A^2 \)). \[ E(t) = \frac{1}{2} k (A(t))^2 = \frac{1}{2} k A_0^2 (e^{-\gamma t})^2 = E_0 e^{-2\gamma t} \] We need to find the time \( t' \) when \( E(t') = E_0/2 \). \[ \frac{E_0}{2} = E_0 e^{-2\gamma t'} \implies e^{-2\gamma t'} = \frac{1}{2} \]
Step 3: Solve for t': From Step 1, we have \( e^{-10\gamma} = 1/2 \). From Step 2, we have \( e^{-2\gamma t'} = 1/2 \). Equating the exponents: \[ -10\gamma = -2\gamma t' \] \[ 10 = 2t' \] \[ t' = 5 \, \text{s} \]