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

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

$\sqrt{\frac{k_{1} + k_{2}}{8 π}}$

$π \sqrt{\frac{k_{1} - k_{2}}{2}}$

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

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

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} Here two spring are in series \\ ∴ Effective spring constant is given by \\ \frac{1}{k} = \frac{1}{k_{1}} + \frac{1}{k_{2}} \\ \Rightarrow k = \frac{k_{1} k_{2}}{k_{1} + k_{2}} \\ ∴ Time period of oscilliation \\ T = 2 π \sqrt{\frac{m}{k}} = 2 π \sqrt{\frac{16}{k_{1} k_{2}} . (k_{1} + k_{2})} \\ = 8 π \sqrt{\frac{k_{1} + k_{2}}{k_{1} k_{2}}} \end{array}$