Question

The angular frequency of the damped oscillator is given by $\omega =\sqrt{(\frac{k}{m}-\frac{r^2}{4m^2})}$ where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio $\frac{r^2}{mk}$ is $8$, the changed in time period compared to the undamped oscillator is approximately as follows:

• A. Increases by 1%
• B. Decreases by 1%
• C. Decreases by 8%
• D. increases by 8%

Answer 1

Answer: (B)

Decreases by 1%

$\omega =\sqrt{\frac{k}{m}-\frac{r^2}{4m^2}}=\sqrt{\frac{k}{m}}\sqrt{1-\frac{r^2}{4mk}}$
$\approx {\omega }_o(1-\frac{r^2}{8mk})\approx$ (1 - 1%)