What is the maximum speed and spring constant of a weight on a massless spring?

[wondermoose]

Homework Statement

A 4.50 kg weight hangs from a vertical massless ideal spring. When set in vertical motion, the weight obeys the equation y(t)=(8.50 cm)sin[(3.85 rad/s)t-1.40]. What is the maximum speed of the weight? What is the spring constant?

Homework Equations

Hooke's law: F = -kx
W(spring): 1/2kx(initial)^2 - 1/2kx(final)^2
W(applied): -W(spring) if stationary before/after displacement

The Attempt at a Solution

Throwing in the equation that the weight obeys has completely thrown me off. I'm not even sure what that means and how I apply it to the problem. I don't have much to show for it because I'm not sure where to begin, and I can't seem to find a similar problem to help find a starting point. Thanks!

 

[juana_sm]

harmonic oscillations are described with this equation: y=y0*sin(omega*t+phi0).
In this case, omega=sqrt(k/m), on the other side, omega = 3.85 rad/s. So you can find k.

as the energy of the system is constant:
k*(y0)^2=m(vmax)^2.
From this equation you can find the maximum velocity.

 

[wondermoose]

Okay, here's what I've got.

omega = sqrt(k/m)

k = omega²*m

k=(3.85 rad/s)²(4.5 kg)
k=66.7 (units?)

Then, k(y₀)² = m(v_max)²

v_max = sqrt((k*y₀²)/m)

v_max = 3.27 m/s

Look okay? I'm not sure how rad²*kg/sec² converts to anything though, in finding the spring constant.

 

[juana_sm]

Yes, it looks fine) about the units of k:
kg/sec^2=(kg*m)/(m*sec^2)=N/m
(N=kg*m/sec^2)