Question

A damped mechanical oscillator of time period 0.5 s has its amplitude reduced to half in 10 oscillations. Calculate (a) damping factor and (b) relaxation time.

• A. The damping factor is $\gamma = \frac{\ln(2)}{5}$ s$^{-1}$ and the relaxation time is $\tau = \frac{5}{\ln(2)}$ s.
• B. The damping factor is $\gamma = \frac{\ln(2)}{10}$ s$^{-1}$ and the relaxation time is $\tau = \frac{10}{\ln(2)}$ s.
• C. The damping factor is $\gamma = \frac{5}{\ln(2)}$ s$^{-1}$ and the relaxation time is $\tau = \frac{\ln(2)}{5}$ s.
• D. The damping factor is $\gamma = \frac{10}{\ln(2)}$ s$^{-1}$ and the relaxation time is $\tau = \frac{\ln(2)}{10}$ s.

Answer 1

Answer: (A)

(a) The damping factor is $\gamma = \frac{\ln(2)}{5}$ s$^{-1}$. (b) The relaxation time is $\tau = \frac{5}{\ln(2)}$ s.

Answer 2

The amplitude of a damped oscillator decays exponentially with time according to the formula $A(t) = A_0 e^{-\gamma t}$. After $n$ oscillations, the time elapsed is $t = nT$, where $T$ is the time period. Thus, the amplitude after $n$ oscillations is $A_n = A_0 e^{-n\gamma T}$.

Given: Time period, $T = 0.5$ s. Amplitude reduces to half in $n = 10$ oscillations, so $A_{10} = \frac{1}{2} A_0$.

Substituting these values into the formula: $\frac{1}{2} A_0 = A_0 e^{-10 \times \gamma \times 0.5}$ $\frac{1}{2} = e^{-5\gamma}$

Taking the natural logarithm of both sides: $\ln\left(\frac{1}{2}\right) = \ln(e^{-5\gamma})$ $-\ln(2) = -5\gamma$ $\gamma = \frac{\ln(2)}{5}$ s$^{-1}$.

The relaxation time, $\tau$, is the time required for the amplitude to decay to $1/e$ of its initial value and is related to the damping factor by $\tau = \frac{1}{\gamma}$.

Using the calculated value of $\gamma$: $\tau = \frac{1}{\frac{\ln(2)}{5}} = \frac{5}{\ln(2)}$ s.

(a) Damping factor $\gamma \approx \frac{0.693}{5} \approx 0.1386$ s$^{-1}$. (b) Relaxation time $\tau \approx \frac{5}{0.693} \approx 7.215$ s.