Asked by JACK

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.
Answered by bobpursley
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
Answered by DoL
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.
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