First slide

Applications of SHM

Question

The time period of simple pendulum is ‘T’ when the length increases by 20cm, its period is $T_{1}$ . When the length is decreased by 20 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

$time period of a simple pendulum is T=2π \sqrt{\frac{l}{g}} here 2π, g are constant T α \sqrt{l} o n s q u a r i n g, k T^{2} = l - - - - - (1) k {T_{1}}^{2} = l + 20 - - - (2) k {T_{2}}^{2} = l - 20 - - - (3) a d d e q u a t i o n (2) a n d (3) k ({T_{1}}^{2} + {T_{2}}^{2}) = 2 l - - - (4) s u b s t i t u t e l f r o m e q u a t i o n (1) k ({T_{1}}^{2} + {T_{2}}^{2}) = 2 (k T^{2}) {T_{1}}^{2} + {T_{2}}^{2} = 2 T^{2}$

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