Question

The time taken for the amplitude of vibrations of a damped oscillator to drop to 25% of its initial value is \( t \). The time taken for its mechanical energy to drop to 50% of its initial mechanical energy is

• A. \( t \)
• B. \( \frac{t}{2} \)
• C. \( \frac{t}{4} \)
• D. \( \frac{t}{8} \)

Hint:

In damped oscillations, the relationship between the amplitude and energy decay is quadratic.

Answer 1

The Correct Option is C

The amplitude of a damped oscillator decays exponentially with time: \[ A(t) = A_0 e^{-\gamma t} \] Where: - \( A_0 \) is the initial amplitude, - \( \gamma \) is the damping coefficient, - \( A(t) \) is the amplitude at time \( t \). The energy of the oscillator is proportional to the square of the amplitude: \[ E(t) \propto A^2(t) \] Therefore, when the amplitude drops to 25% of its initial value, the energy drops to \( 25^2 = 0.0625 \) of its initial value, and when the energy drops to 50%, the amplitude will drop to \( \sqrt{0.5} \) of its initial value. The time taken to reduce the energy by 50% is \( \frac{t}{4} \). Thus, the correct answer is: \[ \boxed{\frac{t}{4}} \]