MCQ
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
  • $2V$
  • B
    $4V$
  • C
    $\frac{V}{2}$
  • D
    $\frac{V}{4}$

Answer

Correct option: A.
$2V$
a
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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