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)

$\omega =\sqrt{\left(\frac{k}{m}-\frac{r^2}{4m^2}\right)}$
${\omega }_0=\sqrt{\frac{k}{m}}$
${\omega }_0-\omega =\sqrt{\frac{k}{m}}-\sqrt{\frac{k}{m}-\frac{r^2}{4m^2}}$
$=\sqrt{\frac{k}{m}}\left(1-\sqrt{1-\frac{r^2}{4mk}}\right)$
$\frac{{\omega }_0-\omega }{{\omega }_0}=\left[1-{\left(1-\frac{r^2}{4mk}\right)}^{1/2}\right]$
$=\left[1-\left(1-\frac{r^2}{8mk}\right)\right]$
$=\frac{r^2}{8mk}=1\%$