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:

In SHM, \[ E\propto A^2 \] So, a small percentage decrease in amplitude produces approximately double the percentage decrease in energy.

Answer 1

The Correct Option is D

Step 1: Recall the relation between energy and amplitude in SHM.
For a simple harmonic oscillator, the total mechanical energy is proportional to the square of the amplitude: \[ E \propto A^2 \] So: \[ \frac{E_2}{E_1} = \left(\frac{A_2}{A_1}\right)^2 \]
Step 2: Find the new amplitude after decrease.
The amplitude decreases by $1.5\%$, so the new amplitude is: \[ A_2 = A_1 - 0.015A_1 = 0.985 A_1 \]
Step 3: Calculate the ratio of new energy to old energy.
\[ \frac{E_2}{E_1} = (0.985)^2 = 0.970225 \]
Step 4: Find the fractional energy lost per cycle.
\[ \frac{\Delta E}{E_1} = 1 - \frac{E_2}{E_1} = 1 - 0.970225 = 0.029775 \] As a percentage: \[ \frac{\Delta E}{E_1} \approx 2.98\% \approx 3\% \]
Step 5: Understand why it is approximately $2 \times 1.5\%$.
Using the approximation $(1 - x)^2 \approx 1 - 2x$ for small $x$: \[ (0.985)^2 \approx 1 - 2(0.015) = 1 - 0.030 \] So energy loss $\approx 3\%$. The energy loss is approximately twice the amplitude loss percentage for small decrements.
Step 6: State the final answer.
The mechanical energy lost by the oscillator in each cycle is approximately $3\%$. \[ \boxed{3\%} \]