The bob of a simple pendulum is displaced from its equilibrium position ‘O’ to a position ‘Q’ which is at a height ‘h’ above ‘O’ and the bob is then released. Assuming the mass of the bob to be ‘m’ and time period of oscillation to be 2.0 sec, the tension in the string when the bob passes through ‘O’ is

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m (g + \frac{π^{2}}{3} h)

m (g + \frac{π^{2}}{2} h)

m (g + 2 π^{2} h)

m (g + π^{2} h)

answer is C.

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Detailed Solution

T is maximum at lowest position

Hence, tension $T = \frac{m υ^{2}}{r} + mg \to (1)$

Here $υ = \sqrt{2gh} \Rightarrow υ^{2} = 2 gh \to (2)$

Given that time period

$∴ T = 2 π \sqrt{\frac{l}{g}} \Rightarrow 2 = 2 π \sqrt{\frac{l}{g}} (∵ T = 2 \sec)$

$g = l π^{2} \Rightarrow l = \frac{g}{π^{2}} \to (3)$

Substitute eq(2) & eq(3) in eq(1)

$T = \frac{m (2gh)}{l} + mg = m \frac{(2gh)}{g} π^{2} + mg$

$T = m (g + 2 π^{2} h)$

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