Physics - EMI/AC Question with Solution | TestHub

The angular frequency of a damped oscillator is given by $\omega =\sqrt{{\omega }_0^2-{\left(\frac{b}{2m}\right)}^2}$. The time period is $T=\frac{2\pi }{\omega }$. For the undamped case, $T_0=\frac{2\pi }{{\omega }_0}$. So, $\frac{T}{T_0}=\frac{{\omega }_0}{\omega }=\frac{{\omega }_0}{\sqrt{{\omega }_0^2-{\left(\frac{b}{2m}\right)}^2}}=\frac{1}{\sqrt{1-\frac{b^2}{4m^2{\omega }_0^2}}}$. For light damping, $\frac{b^2}{4m^2{\omega }_0^2}\ll 1$. Using the binomial approximation $(1-x{)}^{-1/2}\approx 1+\frac{1}{2}x$ for small x:$\frac{T}{T_0}\approx 1+\frac{1}{2}\left(\frac{b^2}{4m^2{\omega }_0^2}\right)=1+\frac{b^2}{8m^2{\omega }_0^2}$. The percentage change in time period is $\left(\frac{T-T_0}{T_0}\right)\times 100\%=\left(\frac{T}{T_0}-1\right)\times 100\%=\frac{b^2}{8m^2{\omega }_0^2}\times 100\%$. Since the term is positive, the time period increases.