## ajprincess 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.

1. amini

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

2. ajprincess

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

3. ajprincess

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

yup.

5. ajprincess

6. ajprincess

where did the 2 pi go ?

8. ajprincess

2pi comes only when we find f right?

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

10. ajprincess

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.

12. ajprincess

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

13. ajprincess

14. rajathsbhat

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.

16. ajprincess

ya it will change.

17. ajprincess

it will be $$\omega'=\omega/2$$.

18. ajprincess

|dw:1354615128244:dw|

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

20. ajprincess

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.

22. ajprincess

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

the k will also change from your first eq.

24. ajprincess

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.

26. ajprincess

so vat I hav done is right?

yup! :)

28. ajprincess

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