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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?

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|

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.

|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 :)