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
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At equilibrium position the speed of a pendulum is equal to $A \omega$. If we double the amplitude, the speed will also double. so the correct option is $'A'$
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