Two identical springs each of spring constant 4k each are attached to a block of mass m and to fixed support. When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is

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$2 π \sqrt{\frac{m}{2 k}}$

$π \sqrt{\frac{2 m}{k}}$

$2 π \sqrt{\frac{m}{k}}$

$\frac{π}{\sqrt{2}} \sqrt{\frac{m}{k}}$

answer is D.

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Detailed Solution

$\begin{array}{l} K_{e f f} = k_{1} + k_{2} = 8 k \\ T i m e p e r i o d T = 2 π \sqrt{\frac{m}{k e f f}} \\ T = 2 π \sqrt{\frac{m}{8 k}} \\ T = \frac{2 π}{2 \sqrt{2}} \sqrt{\frac{m}{k}} \\ T = \frac{π}{\sqrt{2}} \sqrt{\frac{m}{k}} \end{array}$

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