Physics - SHM Question with Solution | TestHub

Time period of simple pendulum For Ist pendulum $\begin{array}{rl}{T}_1& =2\pi \sqrt{\frac{5}{g}}\\ & =2\pi \sqrt{\frac{5}{9.8}}=\sqrt{20}\mathrm{s}\end{array}$ For IInd pendulum $T_2=2\pi \sqrt{\frac{20}{g}}$ $=2\pi \sqrt{\frac{20}{9.8}}=2\sqrt{20}\mathrm{s}$ Suppose the time will be in same phase is $t$ then $\begin{array}{rl}\frac{1}{t}& =\frac{1}{T_1}-\frac{1}{T_2}\\ & =\frac{1}{\sqrt{20}}-\frac{1}{2\sqrt{20}}\end{array}$ $=\frac{1}{2\sqrt{20}}$ So, $t=2\sqrt{20}$ Hence, the number of oscillation of simple pendulum of shorter length $(x)=\frac{2\sqrt{20}}{\sqrt{20}}=2$.