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When a block is suspended by a spring and allowed to oscillate vertically, its time period is 1 sec. When the same block is suspended by another spring, its time period is found to be 2 sec. Now the springs are connected end to end to form a longer spring and the same block is suspended from the combination of springs, the new time period of small vertical oscillations of the block is

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$\sqrt{3} \sec$

$\sqrt{5} \sec$

$\sqrt{\frac{2}{3}} \sec$

3 sec

answer is B.

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Detailed Solution

$T_{1} = 2 π \sqrt{\frac{m}{k_{1}}} \Rightarrow k_{1} = \frac{4 π^{2} m}{{T_{1}}^{2}}; similarly k_{2} = \frac{4 π^{2} m}{{T_{2}}^{2}}$

For series combination of springs,

$\frac{1}{k_{e}} = \frac{1}{k_{1}} + \frac{1}{k_{2}} \Rightarrow \frac{{T_{e}}^{2}}{4 π^{2} m} = \frac{{T_{1}}^{2}}{4 π^{2} m} + \frac{{T_{2}}^{2}}{4 π^{2} m}$

$∴ T_{e} = \sqrt{{T_{1}}^{2} + {T_{2}}^{2}} = \sqrt{1^{2} + 2^{2}} \sec = \sqrt{5} \sec$

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