First slide

Simple hormonic motion

Question

The time period of simple pendulum is ‘T’. When the length increases by 10 cm, its period is T₁. When the length is decreased by
10 cm, its period is T₂. Then the relation between T, T₁ and T₂ is

Easy

A

$\large \frac{2}{{{T^2}}}\, = \,\frac{1}{{T_1^2}} + \frac{1}{{T_2^2}}$
B

$\large \frac{2}{{{T^2}}}\, = \,\frac{1}{{T_1^2}} - \frac{1}{{T_2^2}}$
C

$\large 2{T^2}\, = \,T_1^2 + T_2^2$
D

$\large 2{T^2}\, = \,T_1^2 - T_2^2$

Solution

$\large T\, \propto \,\sqrt l \, \Rightarrow \,l\, = \,k{T^2}.......(1)$
$\large l + 10\, = \,kT_1^2........(2)$
$\large l - 10\, = \,kT_2^2........(3)$
$\large \left( 2 \right) + \left( 3 \right)\, \Rightarrow \,2l\, = \,k\left( {T_1^2 + T_2^2} \right)$
$\large 2k{T^2}\, = \,k\left( {T_1^2 + T_2^2} \right)$
$\large 2{T^2}\, = \,T_1^2 + T_2^2$

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