## anonymous 4 years ago A simple harmonic oscillator with m= 0.560kg and total energy E = 130J has amplitude A= 1.65m . Find the spring constant.

1. anonymous

in order to find k i think we use $w=\sqrt{k/m}$ but we need to find w to find k. I don't know if this is the right way to solve the problem.

2. anonymous

it is w= the square root of k/m

3. anonymous

Hi,i think when mass pass origin all of it's E is EK so$E=(1/2)mv _{2}^{\max}$&$v _{\max}=A \omega$so you can find omeg asd from that you can yield to k

4. anonymous

At the extreme position ,the block has Zero velocity. Total mechanical energy=K.E+P.E=>130=0+1/2KA^2=>K=(260/A^2)=260/(1.65*1.65)=95.5Nm...

5. anonymous

@igbona,if you apply conservation of energy in such a problem then you can easily get your answer. if any query ,tell me.

6. anonymous

so, the x in (1/2)Kx^2 can be substituted by A => (1/2)KA^2. why?

7. anonymous

yea hence you use that formula in amplitude point this point has x=A did you got it?

8. anonymous

yes :)

9. anonymous

because of at extreme position the displacement of the block will be X=A,

10. anonymous

@ igbona,any problem,,

11. anonymous

@Taufique, none at all. I was just curious. Thank you for your help.