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Please help:) A mass less spring hanging vertically, extends by 5cm when a mass M is attached to its lower and vertical oscillations are started by pulling the mass M downwards by another 5cm and then releasing it. i.Find the frequency of oscillation and the maximum velocity of the oscillating mass. ii.If a second mass of 300g is added to M the frequency of oscillation is reduced to half the previous value. Find M.

Physics
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i. Omega= g/0.05 Hertz ii. M=100gr
It wud be really better if u show the steps @amini
|dw:1354605666317:dw||dw:1354605726585:dw| Am I right?

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Other answers:

yup.
thanx. so |dw:1354606338362:dw| right @shadowys?
@Shadowys if u dnt mind pleae check my answer.
where did the 2 pi go ?
2pi comes only when we find f right?
oh. you're not using \(v=\frac{2\pi}{T}\) the answers are different though...
sorry. but \(\omega\) is only equal to (\frac{2\pi}{T}\). Isn't t? Maximum velocity is \(A\omega\)
oh right. sorry. a lil rusty in SHM. right. v=rw so your answer's right.
will the anwr for second part be this? |dw:1354608247839:dw|
i too am a little rusty on SHM but isn't the frequency of oscillation independent of the mass attached to the spring?
wouldn't \(\omega' \)also change because of the increase in mass by 0.3g? @rajathsbhat that is for simple pendulum.
ya it will change.
it will be \(\omega'=\omega/2\).
|dw:1354615128244:dw|
also the i didn't get (M+100) in f'
oops sorry. it has to be M+300.
in g? not in SI units? lol though there wouldn't be much difference anyways. except for the mass part, i think it's okay.
|dw:1354616316357:dw| Am I right nw?
the k will also change from your first eq.
How to find it. New extension is nt given right?
wait. sorry, i was wrong. k is a constant. lol eyes get blurry in a hurry.
so vat I hav done is right?
yup! :)
Thank u sooooooo much. @Shadowys. thanx to u too @rajathsbhat.
you're welcome :)

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