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(D) When a block of mass $m$ is connected to two springs of spring constant $k_1$ and $k_2$ in parallel,the effective spring constant is $k_{eq} = k_1

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$\pi \sqrt{\frac{m}{k}}$

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(D) When a block of mass $m$ is connected to two springs of spring constant $k_1$ and $k_2$ in parallel,the effective spring constant is $k_{eq} = k_1 + k_2$.In this problem,both springs have a spring constant of $2k$.Therefore,the equivalent spring constant is $k_{eq} = 2k + 2k = 4k$.The time period $T$ of a simple harmonic oscillator is given by the formula $T = 2\pi \sqrt{\frac{m}{k_{eq}}}$.Substituting the value of $k_{eq}$,we get $T = 2\pi \sqrt{\frac{m}{4k}}$.Simplifying this,$T = 2\pi \frac{1}{2} \sqrt{\frac{m}{k}} = \pi \sqrt{\frac{m}{k}}$.

Two identical springs of spring constant $2k$ are attached to a block of mass $m$ and to fixed supports (see figure). When the mass is displaced from the equilibrium position on either side,it executes simple harmonic motion. The time period of oscillations of this system is ...... .

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