If the period of oscillation changes from 2 sec to 1.5 sec, and the initial length of simple pendulum is 1m, then the change in length is

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Increased by $\frac{7}{16} m$

decreased by $\frac{7}{16} m$

increased by $\frac{9}{16} m$

decreased by $\frac{9}{16} m$

answer is B.

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Detailed Solution

$\frac{T_{1}}{T_{2}} = \sqrt{\frac{ℓ_{1}}{ℓ_{2}}} = \frac{2}{1 .5} = \sqrt{\frac{ℓ_{1}}{ℓ_{2}}}; ℓ_{2} = \frac{9}{16} m ℓ_{1} = 1 m {Δℓ}_{1} = ℓ_{2} - ℓ_{1} = \frac{9}{16} - 1 = - \frac{7}{16} m$
So, decreases by $\frac{7}{16} m ℓ_{2} < ℓ_{1}$

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