# Spring

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, 8 months ago

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See my answer to this question.

- 4 years, 8 months ago

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

- 4 years, 8 months ago

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

- 4 years, 8 months ago

Yes, as mg=- kx

- 4 years, 8 months ago

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

- 4 years, 8 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..

- 4 years, 8 months ago

Yeah

- 4 years, 8 months ago