What will be elongation in an ideal spring of

  • natural length l,
  • mass m and
  • spring constant k

if it is hung vertically with no mass at the bottom?

Assume the spring obeys Hooke's Law and is sensitive enough to elongate a bit due to gravity.

I think it would be \(\frac{mg}{k}\) assuming it wouldn't make a difference if I take all mass at the bottom.

What do you think?

Note by Lokesh Sharma
4 years, 5 months ago

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See my answer to this question.

A K - 4 years, 5 months ago

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I hope it is 2mg/k,by considering the elongation of centre of mass...right na?

Vinayakraj M - 4 years, 5 months ago

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Yeah it seems plausible but can't be sure as mg/k is as much appealing.

Lokesh Sharma - 4 years, 5 months ago

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Yeah

Harsh Panchal - 4 years, 5 months ago

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Yes, as mg=- kx

Oshin Tiwari - 4 years, 5 months ago

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Why it can't be 2mg/k as suggested by Vinayakraj M above?

Lokesh Sharma - 4 years, 5 months ago

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Yes it should be 2mg/k because the position of COM is twice the original position if u take all the mass at the bottom...he's ri8..

Oshin Tiwari - 4 years, 5 months ago

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