The angular frequency of the damped oscillator is given by, $ω = \sqrt{\frac{k}{m} - \frac{r^{2}}{4 m^{2}}}$ where k is the spring constant, m is the mass of the oscillator and r is the damping constant, If the ratio $\frac{r^{2}}{m k}$ is increased 8%, the change in time period compared to the undamped oscillator is approximately as follows,

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Increases by 1%

Decreases by 1%

Decreases by 8%

Increases by 8%

answer is A.

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Detailed Solution

Angular frequency of undamped oscillator $ω_{0} = \sqrt{\frac{k}{m}}$
Change in angular frequency $Δ ω = - (ω_{i} - ω_{f}) = ω_{0} - ω$

$\Rightarrow Δ ω = - (\sqrt{\frac{k}{m} -} \sqrt{\frac{k}{m} - \frac{r^{2}}{4 m^{2}}})$
$\Rightarrow Δ ω = - \sqrt{\frac{k}{m}} (1 - \sqrt{1 - \frac{r^{2}}{4 m k}})$
$\Rightarrow \frac{Δ ω}{ω_{0}} = - (1 - \sqrt{1 - \frac{0.08}{4}})$
$\Rightarrow \frac{Δ ω}{ω_{0}} = - (1 - \sqrt{1 - 0.02}) = - (1 - (1 - 0.01)) = - 0.01$
$As T = \frac{2 π}{ω}$ $\Rightarrow \frac{Δ T}{T_{0}} = - \frac{Δ ω}{ω_{0}}$
$\Rightarrow$ As $ω$ decreases by 1% $\Rightarrow T$ increases by 1%