Question

The amplitudes of a damped harmonic oscillator after 2 and 4 seconds are \( A_1 \) and \( A_2 \) respectively. If the initial amplitude of the oscillator is \( A_0 \), then

• A. \( A_1 = \sqrt{A_0 A_2} \)
• B. \( A_2 = \sqrt{A_0 A_1} \)
• C. \( A_0 = \sqrt{A_1 A_2} \)
• D. \( A_0 = \frac{A_1 + A_2}{2} \)

Hint:

For a damped oscillator, the ratio of amplitudes at different times can help in deriving relationships between initial and later amplitudes.

Answer 1

The Correct Option is A

In the case of a damped harmonic oscillator, the amplitude decays exponentially over time. The relation between the amplitudes at different times is given by: \[ A(t) = A_0 e^{-\gamma t} \] where \( A_0 \) is the initial amplitude, \( \gamma \) is the damping coefficient, and \( t \) is the time. We are given that the amplitudes after 2 seconds and 4 seconds are \( A_1 \) and \( A_2 \), respectively, and the initial amplitude is \( A_0 \). So, we have: \[ A_1 = A_0 e^{-2\gamma}, \quad A_2 = A_0 e^{-4\gamma} \] Dividing \( A_1 \) by \( A_2 \), we get: \[ \frac{A_1}{A_2} = \frac{A_0 e^{-2\gamma}}{A_0 e^{-4\gamma}} = e^{2\gamma} \] Thus: \[ A_1 = \sqrt{A_0 A_2} \] Therefore, the correct answer is \( A_1 = \sqrt{A_0 A_2} \).

Answer 2

Step 1: Understand damping in harmonic motion.
In a damped harmonic oscillator, the amplitude decreases exponentially with time. The general formula for amplitude at time \( t \) is: \[ A(t) = A_0 e^{-\beta t} \] Where:
- \( A_0 \) is the initial amplitude
- \( \beta \) is the damping coefficient
- \( t \) is time

Step 2: Express amplitudes at 2s and 4s.
\[ A_1 = A(2) = A_0 e^{-2\beta} \] \[ A_2 = A(4) = A_0 e^{-4\beta} \]
Step 3: Solve for \( A_1 \) in terms of \( A_0 \) and \( A_2 \).
From above, \( A_2 = A_0 e^{-4\beta} = (A_0 e^{-2\beta}) \cdot e^{-2\beta} = A_1 \cdot e^{-2\beta} \)
\[ \Rightarrow e^{-2\beta} = \frac{A_2}{A_1} \Rightarrow A_1 = \sqrt{A_0 A_2} \]
Final Answer: \( \boxed{A_1 = \sqrt{A_0 A_2}} \)