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 2 years 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.
 2 years 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.

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amini
 2 years ago
Best ResponseYou've already chosen the best response.0i. Omega= g/0.05 Hertz ii. M=100gr

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0It wud be really better if u show the steps @amini

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1354605666317:dwdw:1354605726585:dw Am I right?

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0thanx. so dw:1354606338362:dw right @shadowys?

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0@Shadowys if u dnt mind pleae check my answer.

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1where did the 2 pi go ?

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.02pi comes only when we find f right?

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

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

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1oh right. sorry. a lil rusty in SHM. right. v=rw so your answer's right.

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0will the anwr for second part be this? dw:1354608247839:dw

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

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

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0it will be \(\omega'=\omega/2\).

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1354615128244:dw

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1also the i didn't get (M+100) in f'

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0oops sorry. it has to be M+300.

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1in 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
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1354616316357:dw Am I right nw?

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1the k will also change from your first eq.

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0How to find it. New extension is nt given right?

Shadowys
 2 years ago
Best ResponseYou've already chosen the best response.1wait. sorry, i was wrong. k is a constant. lol eyes get blurry in a hurry.

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0so vat I hav done is right?

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0Thank u sooooooo much. @Shadowys. thanx to u too @rajathsbhat.
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