Q. The time period of a simple pendulum is $T$. When the length is increased by 10 cm, its period is $T_{1}$. When the length is decreased by 10 cm, its period is $T_{2}$. Then, relation between $T,T_{1}\ and\ T_{2}$ is
A) $\frac{2}{T^{2}} = \frac{1}{T_{1}^{2}} + \frac{1}{T_{2}^{2}}$
B) $\frac{2}{T^{2}} = \frac{1}{T_{1}^{2}} - \frac{1}{T_{2}^{2}}$
C) $2T^{2} = T_{1}^{2} + T_{2}^{2}$
D) $2T^{2} = T_{1}^{2} - T_{2}^{2}$
Explanation
(c) $T^{2} = 4\pi^{2}\left( \frac{l}{g} \right)$ …(i)
$T_{1}^{2} = 4\pi^{2}\left( \frac{l + 10}{g} \right)$ …(ii)
$T_{2}^{2} = 4\pi^{2}\left( \frac{l - 10}{g} \right)\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(iii)$
Adding Eqs. (ii) and (iii), we get
$T_{1}^{2} + T_{2}^{2} = 4\pi^{2}\left\lbrack \frac{2l}{g} \right\rbrack$
$= 2\left( 4\pi^{2} \right)\left( \frac{l}{g} \right) = 2T^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.