Two springs are joined and attached to a mass of 16 kg. The system is then suspended vertically from a rigid support. The spring constant of the two springs are $k_{1}$ and $k_{2}$ respectively. The period of vertical oscillations of the system will be :

8 π \sqrt{\frac{k_{1} + k_{2}}{k_{1} k_{2}}}

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\frac{π}{2} \sqrt{\frac{k_{1}}{k_{2}}}

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\sqrt{\frac{k_{1} + k_{2}}{8 π}}

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\frac{π \sqrt{k_{1} - k_{2}}}{2}

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Solution

The correct option is A $8 π \sqrt{\frac{k_{1} + k_{2}}{k_{1} k_{2}}}$
Here two springs are in series.
$∴$ Effective spring constant is given by $\frac{1}{k} = \frac{1}{k_{1}} + \frac{1}{k_{2}}$
$\Rightarrow$ $k = \frac{k_{1} K_{2}}{k_{1} + k_{2}}$
$∴$ Time period of oscillation
$T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{16}{\frac{k_{1} k_{2}}{(k_{1} + k_{2})}}}$
$= 8 π \sqrt{\frac{k_{1} + k_{2}}{k_{1} k_{2}}}$

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Two springs are joined and attached to a mass of 16 kg. The system is then suspended vertically from a rigid support. The spring constant of the two springs are k1 and k2 respectively. The period of vertical oscillations of the system will be :

The correct option is A 8π√k1+k2k1k2Here two springs are in series.∴ Effective spring constant is given by 1k=1k1+1k2⇒ k=k1K2k1+k2∴ Time period of oscillation T=2π√mk=2π  ⎷16k1k2(k1+k2) =8π√k1+k2k1k2

Two springs are joined and attached to a mass of 16 kg. The system is then suspended vertically from a rigid support. The spring constant of the two springs are $k_{1}$ and $k_{2}$ respectively. The period of vertical oscillations of the system will be :