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(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g}}$,where $\ell$ is the length of the pendulum and $g$ is the accel

Vedclass · Accepted Answer

$\sqrt{2}$

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(A) The time period of a simple pendulum is given by $T = 2 \pi \sqrt{\frac{\ell}{g}}$,where $\ell$ is the length of the pendulum and $g$ is the acceleration due to gravity. The time period is independent of the mass of the bob.Given that the new length $\ell^{\prime} = \frac{\ell}{2}$.The new time period $T^{\prime} = 2 \pi \sqrt{\frac{\ell^{\prime}}{g}} = 2 \pi \sqrt{\frac{\ell}{2g}} = \frac{1}{\sqrt{2}} \left( 2 \pi \sqrt{\frac{\ell}{g}} \right) = \frac{1}{\sqrt{2}} T$.According to the problem,$T^{\prime} = \frac{x}{2} T$.Therefore,$\frac{x}{2} = \frac{1}{\sqrt{2}}$.$x = \frac{2}{\sqrt{2}} = \sqrt{2}$.

If the mass of the bob in a simple pendulum is increased to thrice its original mass and its length is made half its original length,then the new time period of oscillation is $\frac{x}{2}$ times its original time period. Then the value of $x$ is

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