question_answer

The length of a simple pendulum executing SHM is increased by 21%. What is the percentage increase in the time period of the pendulum of increased length.                    

Answer:

                Time period, \[T=2\pi \sqrt{\frac{l}{g}}\] i.e., \[T\propto {{l}^{1/2}}\] The percentage increase in time period is given by                                 \[\frac{\Delta \Tau }{T}\times 100=\frac{1}{2}\frac{\Delta l}{l}\times 100\] \[=\frac{1}{2}\times 21%=\mathbf{10}\mathbf{.5%}\]