A simple pendulum oscillating in air has a period of \(\sqrt{3} \text{s}\). If it is completely immersed in non-viscous liquid, having density \((\frac{1}{4})^{\text{th}}\) of the material of the bob, the new period will be
\(\frac{\sqrt{3}}{2} \text{s}\)
\(\frac{2}{\sqrt{3}} \text{s}\)
Solution:
The effective acceleration due to gravity is \(g' = g\left(1 - \frac{rho_l}{\rho_b}\right) = g\left(1 - \frac{1}{4}\right) = \frac{3}{4}g\). The new time period is \(T' = T\sqrt{\frac{g}{g'}} = \sqrt{3} \times \sqrt{\frac{4}{3}} = 2 \text{s}\).
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