MCQ
When a mass $M$ is attached to the spring of force constant $k$, then the spring stretches by $l$. If the mass oscillates with amplitude $l$, what will be maximum potential energy stored in the spring
  • A
    $\frac{{kl}}{2}$
  • B
    $2kl$
  • $\frac{1}{2}Mgl$
  • D
    $Mgl$

Answer

Correct option: C.
$\frac{1}{2}Mgl$
c
(c) $Mg = Kl$

==> ${U_{\max }} = \frac{1}{2}K{l^2} = \frac{1}{2}mgl$

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