IB Physics SL Simple Harmonics Question

In summary, the problem involves a spring (with a spring constant of 2500N/m) and a 10-kg ball placed on top of the spring. The ball is pushed down an additional 0.5 meters from the equilibrium point and then released. The question asks for the maximum height the ball reaches above its release point. To solve this, we can use conservation of energy instead of simple harmonic motion, since the ball is not attached to the spring.
  • #1
Regalia
1
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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.
 
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  • #2
Hello Regalia,

Welcome to physics forums!

Regalia said:

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. :wink:
 

1. What is Simple Harmonic Motion?

Simple Harmonic Motion (SHM) is a type of periodic motion where an object oscillates back and forth around an equilibrium point with a constant amplitude and period. It occurs when a restoring force is proportional to the displacement from the equilibrium point.

2. How is Simple Harmonic Motion related to IB Physics SL?

Simple Harmonic Motion is a key concept in IB Physics SL and is included in the syllabus under the topic "Mechanics". Students are expected to understand the principles of SHM and be able to apply them to solve problems and analyze real-world situations.

3. How do you calculate the period of a Simple Harmonic Motion?

The period of SHM can be calculated using the formula T = 2π√(m/k), where T is the period in seconds, m is the mass of the object in kilograms, and k is the spring constant in Newtons per meter (N/m).

4. Can you give an example of Simple Harmonic Motion in everyday life?

A common example of SHM is the motion of a pendulum. As the pendulum swings back and forth, it exhibits SHM because the force of gravity acts as a restoring force, bringing the pendulum back to its equilibrium position.

5. How does Simple Harmonic Motion relate to other types of motion?

SHM is a special case of oscillatory motion, which is any type of motion that repeats itself over time. It is also related to circular motion, as the projection of an object moving in a circle onto a straight line results in SHM.

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