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+1 vote
asked in NEET by (1.9k points)

When a mass M is attached to the spring of force constant K,  then the spring stretches  by  l . If the mass oscillates with a amplitude  l. What will be maximum  p.e stored in the spring

1 Answer

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answered by (61.3k points)
edited by

We know that when the mass is suspended, the spring stretches by l, then Mg=kl. And energy stored in spring = kl2/2 = (kl)l/2 = Mgl/2.

When the spring oscillates at the maximum displacement, additional potential energy stored = kl2/2 = Mgl/2.

Therefore maximum potential energy or total potential energy = Mgl/2+Mgl/2 = Mgl

commented by (1.9k points)
Correct Answer   pe= 2kl
commented by (61.3k points)
See, I have updated my answer, it would be Mgl

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