Question

The angular frequency of the damped oscillator is given by,
$\omega = \sqrt{\left(\frac{k}{m} - \frac{r^{2}}{4m^{2}}\right)}$ 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 change in time period compared to the undamped oscillator is approximately as follows :

• A. increases by 1%
• B. increases by 8%
• C. decreases by 1%
• D. decreases by 8%

Answer 1

Answer: (A)

$\omega = \sqrt{\left(\frac{k}{m} - \frac{r^{2}}{4m^{2}}\right)}$
$\omega_{0} = \sqrt{\frac{k}{m} } $
$\omega _{0} -\omega=\sqrt{\frac{k}{m} }-\sqrt{\frac{k}{m}-\frac{r^{2}}{4m^{2}}} $
$= \sqrt{\frac{k}{m} }\left(1-\sqrt{1-\frac{r^{2}}{4mk}}\right)$
$\frac{\omega _{0} -\omega }{\omega _{0} }= \left[1-\left(1-\frac{r^{2}}{4mk}\right)^{1/2}\right]$
$= \left[1-\left(1-\frac{r^{2}}{8mk}\right)\right]$
$= \frac{r^{2}}{8mk} = 1\%$