Question:

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 :

Updated On: Dec 4, 2024
  • \(\sqrt3\)
  • \(\sqrt2\)
  • \(2\sqrt3\)
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The Correct Option is B

Solution and Explanation

The time period of a pendulum is:
$T = 2\pi\sqrt{\frac{L}{g}}$.
If the length $L$ is halved, the new time period becomes:
$T_new = 2\pi\sqrt{\frac{L/2}{g}} = 2\pi \cdot \frac{1}{\sqrt{2}}\sqrt{\frac{L}{g}} = \frac{T}{\sqrt{2}}$.
Comparing with $\frac{x}{2}$:
$\frac{T}{\sqrt{2}} = \frac{x}{2}T \implies x = \sqrt{2}$.

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