IB Physics SL Simple Harmonics Question

[Regalia]

Homework Statement

A spring (spring constant=2500N/m) is fixed vertically to the floor. A 10-kg ball is placed on the spring, pushed down 0.5m, and released. How high does the ball fly above its release position?

Homework Equations

x=Xo cos wt
w= √(k/m)
T=2pi√(m/k)

The Attempt at a Solution

x=Xo cos(15.8)(0.397)

I have solved for the frequency and the period, but I don't know what values I should plug in for Xo and x.

That is as far as I could get, I have no idea how to solve for the amplitude (Xo), which appears to be the answer to the question. I'm pretty slow on physics, so sorry if this question seems elementary/stupid.

Thanks.

edit: I have not learned calculus yet, if that makes a difference when solving this question.

 

[collinsmark]

Hello Regalia,

Welcome to physics forums!

Homework Statement

A spring (spring constant=2500N/m) is fixed vertically to the floor. A 10-kg ball is placed on the spring, pushed down 0.5m, and released. How high does the ball fly above its release position?

The problem statement is somewhat vague on the setup. I'm going to assume that the ball is placed on the spring, and allowed to reach a state of equilibrium. Then the ball is pushed down an additional 0.5 meters from the equilibrium point.

(As opposed to the ball being pushed down 0.5 meters from the spring's zero compression point.)

Homework Equations

x=Xo cos wt
w= √(k/m)
T=2pi√(m/k)

The Attempt at a Solution

x=Xo cos(15.8)(0.397)

I have solved for the frequency and the period, but I don't know what values I should plug in for Xo and x.

That is as far as I could get, I have no idea how to solve for the amplitude (Xo), which appears to be the answer to the question. I'm pretty slow on physics, so sorry if this question seems elementary/stupid.

I interpret the problem statement such that the ball is not physically attached/connected to the spring. It is simply resting on top. That makes a big difference for this problem.

If the ball were attached/connected to the top of the spring, the amplitude of the oscillation would simply be 0.5 meters. Once the ball is released, the spring would push it up to the equilibrium point, where the ball reaches its maximum velocity. Then the spring would pull back on the ball as it continues to rise. Soon, the ball would rise up a maximum of 1.0 meters above the release point (which is 0.5 m above the equilibrium point), where spring finally wins out and pulls it back down. And ignoring friction, the process would continue indefinitely. That's simple harmonic motion.

But, at least as I interpret the problem statement, the ball is not physically attached to the top of the spring. The ball is free to shoot off the top of the spring, once the ball reaches the equilibrium point. Simple harmonic motion doesn't apply here.

Hint: Try conservation of energy instead.