Physics
posted by Dan on .
A particle of mass m = 0.40 kg oscillates on the end of a spring with an amplitude of 0.30 m. At t = 0, it's at it's equilibrium position (x=0) and is moving in the positive xdirection. At 0.20 s later it is at x = 0.25 m.
a. What is the lowest possible value for the spring constant?
b. What is the second lowest possible value for the spring constant?

if 0 at 0 and v+ then
x = .3 sin 2 pi f t
.25 = .3 sin 2 pi f(.2)
.8333 = sin 2 pi f(.2)
so
2 pi f(.2) = sin^1 .8333 = 56.4 deg = .985 radians
so
2 pi f = 4.92 radians/second
but
w = sqrt(k/m) = 2 pi f
so
k/m = (4.92)^2
k = .4 (4.92)^2
the next one will be at twice the frequency so it makes a second trip in that time
2 pi f = w = 9.84 radians /second 
Thanks, I wasn't able to get the second part of the answer (got a spring constant of 24.21 N/m) correct though.

For the second part, since this is a sine function, there's a particular angle that would have the same value as 0.8333. Like sin(0 deg) and sin(180 deg) equals to 0. So I did 180  56.4 to get that particular angle. Once I got that, I converted it to radians and repeated the steps you provided after that. Thanks for the help Damon.