Question

If the amplitude of a lightly damped oscillator decreases by $ 1.5% $, then the mechanical energy of the oscillator lost in each cycle is:

• A. \( 1.5\% \)
• B. \( 0.75\% \)
• C. \( 6\% \)
• D. \( 3\% \)

Hint:

When amplitude drops by small percentage \( x\% \), energy drops by approximately \( 2x\% \) since \( E \propto A^2 \).

Answer 1

The Correct Option is D

The energy of a harmonic oscillator is proportional to the square of the amplitude: \[ E \propto A^2 \] If amplitude decreases by \( 1.5% \), then: \[ \frac{A'}{A} = 1 - \frac{1.5}{100} = 0.985 \Rightarrow \left( \frac{A'}{A} \right)^2 = 0.985^2 \approx 0.9702 \] So, percentage energy lost: \[ \Delta E = (1 - 0.9702) \times 100 \approx 2.98% \approx 3% \]