Question 114 of 210

A mass $m$ is attached to two springs as shown in figure. The spring constants of two springs are ${K}_{1}$ and ${K}_{2}.$ For the frictionless surface, the time period of oscillation of mass $m$ is

A$2\pi \sqrt{\dfrac{m}{{K}_{1}+{K}_{2}}}$B$\dfrac{1}{2\pi }\sqrt{\dfrac{{K}_{1}-{K}_{2}}{m}}$C$2\pi \sqrt{\dfrac{m}{{K}_{1}-{K}_{2}}}$D$\dfrac{1}{2\pi }\sqrt{\dfrac{{K}_{1}+{K}_{2}}{m}}$

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Answer: (A) $2\pi \sqrt{\dfrac{m}{{K}_{1}+{K}_{2}}}$

For the given combination, if the block is displaced towards right, elongation in one spring will be equal to the compression in other and the direction of force due to both spring will be the same and restoring in nature. Hence, both springs are effectively in parallel combination.
Therefore,
$\Rightarrow {K}_{eff}={K}_{1}+{K}_{2}$ and
$\Rightarrow T=2\pi \sqrt{\dfrac{m}{{K}_{eff}}}=2\pi \sqrt{\dfrac{m}{{K}_{1}+{K}_{2}}}$