Question

45.The timeperiodof oscillation of a simple pendulum is square root 2 sec.If its length is decreased to half of initial length, then its new period is -----------

Answer 1

$$\begin{gathered} \mathbf{Time}\mathbf{}\mathbf{period}\mathbf{}\mathbf{of}\mathbf{}\mathbf{a}\mathbf{}\mathbf{pendulum}\mathbf{:} \\ \mathrm{T}\mathrm{\alpha }\sqrt{\mathrm{L}} \\ \frac{\mathrm{T}\text{'}}{\mathrm{T}}=\sqrt{\frac{\mathrm{L}\text{'}}{\mathrm{L}}} \\ \mathrm{T}\text{'}=2\times \sqrt{\frac{\mathrm{L}/2}{\mathrm{L}}} \\ =\frac{2}{\sqrt{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\sqrt{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{s} \end{gathered}$$