A simple pendulum is executing simple harmonic motion with a time period T . If length of pendulum i

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13.Oscillations

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A simple pendulum is executing simple harmonic motion with a time period $T$. If the length of the pendulum is increased by $21\%$, the percentage increase in the time period of the pendulum of increased length is ..... $\%$

A

$10$

B

$21$

C

$30$

D

$50$

(AIEEE-2003) (AIIMS-2001)

Solution

(a) If initial length ${l_1} = 100$ then ${l_2} = 121$

By using $T = 2\pi \sqrt {\frac{l}{g}} $

$\Rightarrow \frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{{l_1}}}{{{l_2}}}} $

Hence, $\frac{{{T_1}}}{{{T_2}}} = \sqrt {\frac{{100}}{{121}}} $

$\Rightarrow {T_2} = 1.1\,{T_1}$

$\%$ increase = $\frac{{{T_2} – {T_1}}}{{{T_1}}} \times 100 = 10\,\% $

Standard 11

Physics

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