- ajprincess

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.

- schrodinger

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- anonymous

i. Omega= g/0.05 Hertz
ii. M=100gr

- ajprincess

It wud be really better if u show the steps @amini

- ajprincess

|dw:1354605666317:dw||dw:1354605726585:dw|
Am I right?

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## More answers

- anonymous

yup.

- ajprincess

thanx. so |dw:1354606338362:dw| right @shadowys?

- ajprincess

@Shadowys if u dnt mind pleae check my answer.

- anonymous

where did the 2 pi go ?

- ajprincess

2pi comes only when we find f right?

- anonymous

oh. you're not using \(v=\frac{2\pi}{T}\) the answers are different though...

- ajprincess

sorry. but \(\omega\) is only equal to (\frac{2\pi}{T}\). Isn't t? Maximum velocity is \(A\omega\)

- anonymous

oh right. sorry. a lil rusty in SHM. right. v=rw so your answer's right.

- ajprincess

will the anwr for second part be this?
|dw:1354608247839:dw|

- ajprincess

- anonymous

i too am a little rusty on SHM but isn't the frequency of oscillation independent of the mass attached to the spring?

- anonymous

wouldn't \(\omega' \)also change because of the increase in mass by 0.3g?
@rajathsbhat that is for simple pendulum.

- ajprincess

ya it will change.

- ajprincess

it will be \(\omega'=\omega/2\).

- ajprincess

|dw:1354615128244:dw|

- anonymous

also the i didn't get (M+100) in f'

- ajprincess

oops sorry. it has to be M+300.

- anonymous

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.

- ajprincess

|dw:1354616316357:dw|
Am I right nw?

- anonymous

the k will also change from your first eq.

- ajprincess

How to find it. New extension is nt given right?

- anonymous

wait. sorry, i was wrong. k is a constant. lol eyes get blurry in a hurry.

- ajprincess

so vat I hav done is right?

- anonymous

yup! :)

- ajprincess

Thank u sooooooo much. @Shadowys. thanx to u too @rajathsbhat.

- anonymous

you're welcome :)

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