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 solve this problem, we need to understand the behavior of a damped harmonic oscillator. The amplitude of such an oscillator decreases exponentially over time. The relationship can be expressed as:

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

where:

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

Given that the amplitude reduces to half in 10 seconds, we can write the equation as:

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

Solving for \( \lambda \):

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

Taking the natural logarithm on both sides:

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

This simplifies to:

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

Now, we find the amplitude after 30 seconds. The equation becomes:

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

Substituting \( \lambda \):

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

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

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

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

Answer 2

Amplitude decays exponentially in damped SHM: \[ A(t) = A_0 e^{-bt} \] Given: \[ A(10) = \frac{A_0}{2} = A_0 e^{-10b} \Rightarrow e^{-10b} = \frac{1}{2} \] Take log: \[ -10b = \ln\left(\frac{1}{2}\right) = -\ln 2 \Rightarrow b = \frac{\ln 2}{10} \]  Find \( A(30) \): \[ A(30) = A_0 e^{-30b} = A_0 e^{-3\ln 2} = A_0 \cdot 2^{-3} = \frac{A_0}{8} \]