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Q.

Calculate the change in the length of a simple pendulum of length 1m required to change its period of oscillation changes from 2s to 1.5s.

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Detailed Solution

$\begin{array}{l} \frac{T_{1}}{T_{2}} = \sqrt{\frac{ℓ_{1}}{ℓ_{2}}}; \frac{2}{1 .5} = \sqrt{\frac{ℓ_{1}}{ℓ_{2}}} \\ \frac{4}{3} = \sqrt{\frac{1}{l^{2}}} \Rightarrow \frac{16}{9} = \frac{1}{l^{2}} \Rightarrow l^{2} = \frac{9}{16} \end{array}$
Change in length $Δℓ = ℓ_{2} - ℓ_{1}$
$Δℓ = \frac{9}{16} - 1 Δℓ = \frac{- 7}{16}$
$∴ Decreases by \frac{7}{16} m .$

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