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?

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TopNewestSee my answer to this question. – A K · 2 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 · 2 years, 5 months ago

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– Lokesh Sharma · 2 years, 5 months ago

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

Yeah – Harsh Panchal · 2 years, 5 months ago

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Yes, as mg=- kx – Oshin Tiwari · 2 years, 5 months ago

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– Lokesh Sharma · 2 years, 5 months ago

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

– Oshin Tiwari · 2 years, 5 months ago

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..Log in to reply