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: Understanding the Concept:
In a damped harmonic oscillation, both the amplitude and the total mechanical energy decrease exponentially with time. The key relationship is that the mechanical energy is proportional to the square of the amplitude (\(E \propto A^2\)).
Step 2: Key Formulae:
The amplitude of a damped oscillator at time \(t\) is given by:
\[ A(t) = A_0 e^{-\gamma t} \] where \(A_0\) is the initial amplitude and \(\gamma\) is a damping constant.
The total mechanical energy of the oscillator at time \(t\) is:
\[ E(t) = \frac{1}{2} k A(t)^2 = \frac{1}{2} k (A_0 e^{-\gamma t})^2 = \left(\frac{1}{2} k A_0^2\right) e^{-2\gamma t} \] So, we can write:
\[ E(t) = E_0 e^{-2\gamma t} \] where \(E_0\) is the initial energy.
Step 3: Detailed Explanation:
Using the Amplitude Information:
We are given that the amplitude becomes half its initial value in 10 s. So, at \(t = 10\) s, \(A(10) = A_0/2\).
\[ \frac{A_0}{2} = A_0 e^{-\gamma(10)} \] \[ \frac{1}{2} = e^{-10\gamma} \quad \cdots(1) \] Finding the Time for Energy to Halve:
We need to find the time \(t'\) when the energy becomes half its initial value. So, \(E(t') = E_0/2\).
\[ \frac{E_0}{2} = E_0 e^{-2\gamma t'} \] \[ \frac{1}{2} = e^{-2\gamma t'} \quad \cdots(2) \] Comparing the two expressions:
From equation (1) and (2), we can see that the expressions on the right-hand side are equal. Therefore, their exponents must be equal.
\[ e^{-10\gamma} = e^{-2\gamma t'} \] \[ -10\gamma = -2\gamma t' \] Assuming \(\gamma \neq 0\), we can cancel \(- \gamma\) from both sides.
\[ 10 = 2t' \] \[ t' = \frac{10}{2} = 5 \text{ s} \] Step 4: Final Answer:
The time taken for the mechanical energy to become half of its initial value is 5 s. Therefore, option (D) is correct.