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Physics · 2014

JEE Main · 11 April 2014, Shift 1 · Q14

The angular frequency of the damped oscillator is given by, ω=√((k/m-(r^2)/(4 m^2))) where k is the spring constant, m is the mass of the oscillator…

The angular frequency of the damped oscillator is given by, $\displaystyle \omega=\sqrt{\left(\frac{\mathrm{k}}{\mathrm{m}}-\frac{\mathrm{r}^2}{4 \mathrm{~m}^2}\right)}$ where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio $\displaystyle \frac{\mathrm{r}^2}{\mathrm{mk}}$ is $\displaystyle 8 \%$, the change in time period compared to the undamped oscillator is approximately as follows :
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JEE Main 2014 Physics question, with the answer from NTA’s final answer key. Where our answers come from.