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A body of mass M, executes vertical SHM with periods 3 sec and 4 sec, when separately attached to spring A and spring B respectively. The period of SHM, when the body executes SHM, as shown in the figure is $t_{0}$ . Then

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$5 s e c$

2.4 sec

3.24 sec

5.64 sec

answer is B.

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Detailed Solution

$t = 2 π \sqrt{\frac{m}{k}}$

$\Rightarrow k = Const. t^{- 2}$

Here the springs are joined in parallel. So

$k_{0} = k_{1} + k_{2}$

where k₀ is resultant force constant

$∴ Const. t_{0}^{- 2} = {Const.t}_{1}^{- 2} + {Const.t}_{2}^{- 2}$

$Or, t_{0}^{- 2} = t_{1}^{- 2} + t_{2}^{- 2}$ $t_{0} = \frac{t_{1} t_{2}}{\sqrt{{(t_{1})}^{2} + {(t_{2})}^{2}}} = \frac{3 \times 4}{\sqrt{3^{2} + 4^{2}}} s e c = 2.4 s e c$

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