Question

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

Answer 1

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.

Answer 2

it is w= the square root of k/m

Answer 3

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

Answer 4

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...

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

yes :)

Answer 9

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

Answer 10

@ igbona,any problem,,

Answer 11

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