Physics - SHM Question with Solution | TestHub

Given, $T=0.5\mathrm{s}$, We know that, $T=2\pi \sqrt{\frac{m}{k}}\Rightarrow 0.5=2\pi \sqrt{\frac{m}{k}}$ $\frac{1}{4\pi }=\sqrt{\frac{m}{k}}$ On squaring both sides, we get $\frac{m}{k}=\frac{1}{16{\pi }^2}$...(i) $\begin{array}{rl}& \Rightarrow T+0.1=2\pi \sqrt{\frac{m+m^{\prime }}{k}}\\ & \Rightarrow 0.5+0.1=2\pi \sqrt{\frac{m+m^{\prime }}{k}}\\ & \Rightarrow {\left(\frac{0.6}{2\pi }\right)}^2=\frac{m+m^{\prime }}{k}=\frac{m}{k}+\frac{m^{\prime }}{k}\\ & \end{array}$ $\Rightarrow {\left(\frac{0.6}{2\pi }\right)}^2=\frac{1}{16{\pi }^2}+\frac{m^{\prime }}{k}$ [using Eq. (i)] $\Rightarrow \frac{(0.6{)}^2}{4{\pi }^2}-\frac{1}{16{\pi }^2}=\frac{m^{\prime }}{k}$ $\Rightarrow \frac{m^{\prime }}{k}=\frac{1.44-1}{16{\pi }^2}\Rightarrow \frac{m^{\prime }}{k}=\frac{0.44}{16{\pi }^2}$....(ii) $\begin{array}{rlr}\because & m^{\prime }g& =kx\\ \therefore & x& =\frac{m^{\prime }}{k}g\end{array}$ $=\frac{0.44}{16{\pi }^2}\times 10$ [using Eq. (ii)] $\begin{array}{rl}& =0.0279\\ & =2.79\times 10^{-2}\mathrm{m}\\ & =2.7\mathrm{cm}\end{array}$