Question

A 0.181 kg mass is attached to a spring and executes simple harmonic motion with a pe-riod of 0.21 s. The total energy of the system is 1.2 J.
Find the force constant of the spring.

b) Find the amplitude of the motion.

Answer 1

total energy
1.2J
Well, when amplitude is max, it is all PEnergy, no KE.

f(t)=Asin(2PI/.21 * t)
v(t)=-A2PI/.21 * cos (2PIt/.21)
but when cos is =0, velocity is zero, and all PE is in spring
.21=1/2 kx^2 where x= A
solve for k

Answer 2

bobpursley, you just subbed in time for energy. I actually have the same question as JACK. and what you're saying doesn't make much sense to me. Also, we can't solve for k using that equation without x or A which we are looking for in part b.