First slide

Applications of SHM

Question

The time period of simple pendulum is ‘T’ when the length increases by 10cm, its period is $T_{1}$ . When the length is decreased by 10 cm, its time period is $T_{2}$ . When the relation between T, $T_{1}$ and $T_{2}$ is

Moderate

A

$\frac{2}{T^{2}} = \frac{1}{{T_{1}}^{2}} + \frac{1}{{T_{2}}^{2}}$
B

$\frac{2}{T^{2}} = \frac{1}{{T_{1}}^{2}} - \frac{1}{{T_{2}}^{2}}$
C

$2 T^{2} = {T_{1}}^{2} + {T_{2}}^{2}$
D

$2 T^{2} = {T_{1}}^{2} - {T_{2}}^{2}$

Solution

$t i m e p e r i o d o f s i m p l e p e n d u l u m i s T = 2 π \sqrt{\frac{l}{g}}$
$Tα \sqrt{l}$ other terms are constants
${l=KT}^{2}$ -----(1) proportionality constant is K,
by the question,
${l+10=KT}_{1}^{2}$ ----(2)
${l-10=KT}_{2}^{2}$ -----(3)
Adding (2) and (3) equations
$2 l = K ({T_{1}}^{2} {+T}_{2}^{2})$
substitute (1)
$2 ({KT}^{2}) = K ({T_{1}}^{2} {+T}_{2}^{2})$
$2 T^{2} = (T_{1}^{2} + T_{2}^{2})$

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