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?

#PhysicsProblem #Springs

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Comments

See my answer to this question.

A K - 7 years, 1 month ago

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

Vinayakraj M - 7 years, 1 month ago

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

Lokesh Sharma - 7 years, 1 month ago

Yes, as mg=- kx

Oshin Tiwari - 7 years, 1 month ago

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

Lokesh Sharma - 7 years, 1 month 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..

Oshin Tiwari - 7 years, 1 month ago

Yeah

harsh panchal - 7 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$