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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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aminiBest ResponseYou've already chosen the best response.0
i. Omega= g/0.05 Hertz ii. M=100gr
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
It wud be really better if u show the steps @amini
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
dw:1354605666317:dwdw:1354605726585:dw Am I right?
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
thanx. so dw:1354606338362:dw right @shadowys?
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
@Shadowys if u dnt mind pleae check my answer.
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
where did the 2 pi go ?
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
2pi comes only when we find f right?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
oh. you're not using \(v=\frac{2\pi}{T}\) the answers are different though...
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
sorry. but \(\omega\) is only equal to (\frac{2\pi}{T}\). Isn't t? Maximum velocity is \(A\omega\)
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
oh right. sorry. a lil rusty in SHM. right. v=rw so your answer's right.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
will the anwr for second part be this? dw:1354608247839:dw
 one year ago

rajathsbhatBest ResponseYou've already chosen the best response.0
i too am a little rusty on SHM but isn't the frequency of oscillation independent of the mass attached to the spring?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
wouldn't \(\omega' \)also change because of the increase in mass by 0.3g? @rajathsbhat that is for simple pendulum.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
it will be \(\omega'=\omega/2\).
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
dw:1354615128244:dw
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
also the i didn't get (M+100) in f'
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
oops sorry. it has to be M+300.
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
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.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
dw:1354616316357:dw Am I right nw?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
the k will also change from your first eq.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
How to find it. New extension is nt given right?
 one year ago

ShadowysBest ResponseYou've already chosen the best response.1
wait. sorry, i was wrong. k is a constant. lol eyes get blurry in a hurry.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
so vat I hav done is right?
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
Thank u sooooooo much. @Shadowys. thanx to u too @rajathsbhat.
 one year ago
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