Simple hormonic motion

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Question

A 2 kg block is suspended by a spring and for small vertical oscillations, time period is found to be 1 sec. Now another mass m is attached to the 2kg block and this time the new time period of oscillation is found to be 1.5 sec. Then the value of m (in kg) is

Moderate

Solution

$T_{1} = 2 π \sqrt{\frac{M}{K}} and T_{2} = 2 π \sqrt{\frac{m + M}{K}}$
$∴ \frac{T_{2}}{T_{1}} = \sqrt{\frac{m + M}{M}}$
$\Rightarrow m = M [{(\frac{T_{2}}{T_{1}})}^{2} - 1] = 2 (\frac{9}{4} - 1) = 2.5 kg$

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