Question

A damped harmonic oscillator has an amplitude that reduces to half in 10 seconds. What will be the amplitude after 30 seconds?

• A. \( \frac{1}{4} \) of original amplitude
• B. \( \frac{1}{8} \) of original amplitude
• C. \( \frac{1}{16} \) of original amplitude
• D. \( \frac{1}{2} \) of original amplitude

Hint:

\textbf{Tip:} In damped SHM, use \( A(t) = A_0 e^{-bt} \); halving time gives exponential decay constant.

Answer 1

The Correct Option is B

To address this issue, we must first comprehend the behavior of a damped harmonic oscillator. The amplitude of such an oscillator diminishes exponentially over time. This relationship is defined by the equation:

\( A(t) = A_0 e^{-\lambda t} \)

In this formula:

• \( A(t) \) represents the amplitude at time \( t \).
• \( A_0 \) is the initial amplitude.
• \( \lambda \) denotes the damping coefficient.
• \( e \) is the base of the natural logarithm.

Given that the amplitude halves within 10 seconds, we can formulate the equation as:

\( \frac{1}{2}A_0 = A_0 e^{-\lambda \times 10} \)

To solve for \( \lambda \):

\( \frac{1}{2} = e^{-10\lambda} \)

Applying the natural logarithm to both sides yields:

\( \ln\left(\frac{1}{2}\right) = -10\lambda \)

This equation simplifies to:

\( \lambda = -\frac{\ln\left(\frac{1}{2}\right)}{10} = \frac{\ln(2)}{10} \)

Now, we will determine the amplitude at 30 seconds. The governing equation is:

\( A(30) = A_0 e^{-\lambda \times 30} \)

Substituting the value of \( \lambda \):

\( A(30) = A_0 e^{-\frac{30 \ln(2)}{10}} = A_0 e^{-3 \ln(2)} \)

Recognizing that \( e^{-\ln(2)} = \frac{1}{2} \), it follows that \( e^{-3 \ln(2)} = \left(\frac{1}{2}\right)^3 \):

\( A(30) = A_0 \left(\frac{1}{2}\right)^3 = A_0 \times \frac{1}{8} \)

Consequently, after 30 seconds, the amplitude will be \( \frac{1}{8} \) of its original value.