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Question: Answered & Verified by Expert
The time period of a simple pendulum is $T$. When the length is increased by $10 \mathrm{~cm}$, its period is $T_1$. When the length is decreased by $10 \mathrm{~cm}$, its period is $T_2$. Then, relation between $T, T_1$ and $T_2$ is
PhysicsOscillationsTS EAMCETTS EAMCET 2004
Options:
  • 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 $2 T^2=T_1^2+T_2^2$
  • D $2 T^2=T_1^2-T_2^2$
Solution:
1227 Upvotes Verified Answer
The correct answer is: $2 T^2=T_1^2+T_2^2$


Adding Eqs. (ii) and (iii), we get
$\begin{aligned} T_1^2+T_2^2 & =4 \pi^2\left[\frac{2 l}{g}\right] \\ & =2\left(4 \pi^2\right)\left(\frac{l}{g}\right)=2 T^2 .\end{aligned}$

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