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Variation of G Due to Height and Depth

The time peri...

Question

The time period of simple pendulum is T. When the length increases by 10 cm, its period is $T_{1}$ . When the length is decreased by 10 cm, its period is $T_{2}$ .Then the relation between T, $T_{1}$ and $T_{2}$ is

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Solution

The correct option is C

$T = 2 π \sqrt{\frac{l}{g}}$

$\Rightarrow T_{1} = 2 π \sqrt{\frac{l + 10}{g}}$

$\Rightarrow l + 10 = \frac{g T_{1}^{2}}{4 π^{2}} (i)$

$T_{2} = 2 π \sqrt{\frac{l - 10}{g}} \Rightarrow l - 10 = \frac{g T_{2}^{2}}{4 π^{2}} \to (i i)$

Adding (i) and (ii) we get

$l = \frac{\frac{g T_{1}^{2}}{4 π^{2}} + \frac{g T_{2}^{2}}{4 π^{2}}}{2} = \frac{g T^{2}}{4 π^{2}}$

$\Rightarrow 2 T^{2} = T_{1}^{2} + T_{2}^{2}$

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