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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