A mass at the end of a spring executes harmonic motion about an equilibrium position with an amplitude $A.$ Its speed as it passes through the equilibrium position is $V.$ If extended $2A$ and released, the speed of the mass passing through the equilibrium position will be

  • A

    $2V$

  • B

    $4V$

  • C

    $\frac{V}{2}$

  • D

    $\frac{V}{4}$

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