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Normal Reaction

A body in SHM...

Question

A body in SHM has time period $T_{1}$ seconds under influence of one force and the same body executes SHM with period $T_{2}$ seconds under the influence of another force. Find the time period (in s) of the same body when both forces act simultaneously in the same direction.

A

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B

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C

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D

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Solution

The correct option is D
$F_{1} = \frac{- 4 π^{2}}{T_{1}^{2}} x$
$F_{2} = \frac{- 4 π^{2}}{T_{2}^{2}} x$
$\Rightarrow F_{1} + F_{2} = - 4 π^{2} (\frac{1}{T_{1}^{2}} + \frac{1}{T_{2}^{2}}) x$
$= \frac{- 4 π^{2}}{T^{2}} x$
$∴ T = \sqrt{\frac{T_{1}^{2} T_{2}^{2}}{T_{1}^{2} + T_{2}^{2}}}$

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