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)

 

[rwisz]

I would first clarify the given information and assumptions. It is stated that the weight is hanging from a vertical massless ideal spring, which means that the spring has no mass and is perfectly elastic. This also implies that the weight is moving only in the vertical direction and there are no external forces acting on it.

Using Hooke's law, we can write the equation of motion for the weight as F = -kx = ma, where F is the force exerted by the spring, k is the spring constant, x is the displacement from the equilibrium position, m is the mass of the weight, and a is the acceleration.

We can also use the given equation y(t) = (8.50 cm)sin[(3.85 rad/s)t-1.40] to find the displacement of the weight at any given time t. Since the maximum displacement of the weight is 8.50 cm, we can set this equal to x and solve for t to find the time at which the weight reaches its maximum displacement.

Once we have the time, we can use the equation of motion to find the acceleration of the weight at that time. From there, we can use the equation v = u + at to find the maximum speed of the weight, where u is the initial velocity (which is 0 since the weight starts from rest).

To find the spring constant, we can use the equation W(spring) = 1/2kx(initial)^2 - 1/2kx(final)^2, where W(spring) is the work done by the spring, x(initial) is the initial displacement (in this case, 0), and x(final) is the final displacement (8.50 cm). Since the work done by the spring is equal to the work done by the weight, we can set W(applied) = -W(spring) and solve for k.

I hope this helps to get you started on solving the problem. Remember to always clarify the given information and assumptions, and to use the appropriate equations and concepts to solve the problem.