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Simple Pendulum

In an experim...

Question

In an experiment to determine the period of a simple pendulum of length $1 m,$ it is attached to different spherical bobs of radii $r_{1}$ and $r_{2}$ . The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be $5 \times 10^{- 4} s$ , the difference in radii, $| r_{1} - r_{2} |$ is best given by :

A

0.01 c m

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B

0.5 c m

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C

1 c m

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D

0.1 c m

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Solution

The correct option is C $0.1 c m$
$T_{1} = \sqrt{\frac{2 ({l + r}_{1})}{g}}$
$T_{2} = \sqrt{\frac{2 (l + r_{2})}{g}}$
$\frac{T_{1} - T_{2}}{T_{1}} = \frac{\sqrt{2 (l + r_{1})} - \sqrt{2 (l + r_{2})}}{\sqrt{2 (l + r_{1})}}$

Upon simplification,

$5 \times 10^{- 4} = \frac{\sqrt{2 (l + r_{1})} - \sqrt{2 (l + r_{2})}}{\sqrt{2 (l + r_{1})}}$
$\sqrt{l + r_{1}} - \sqrt{l + r_{2}} = (5 \times 10^{- 4}) \times \sqrt{l + r_{1}} \sqrt{l + r_{2}} = (1 - 5 \times 10^{- 4}) (\sqrt{l + r_{1}}) l + r_{2} = (1 - 5 \times 10^{- 4})^{2} (l + r_{1}) l + r_{2} = 1 + 2 \times 5 \times 10^{- 4} (l + r_{1}) l + r_{2} = 1 + 10 \times 10^{- 4} (l + r_{1}) l + r_{2} = 1.001 (l + r_{1}) r_{2} - r_{1} = 1 - 1.001 r_{2} - r_{1} = 0.001 | r_{2} - r_{1} | = 0.001 m | r_{2} - r_{1} | = 0.1 c m$

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