# Homework Help: Conceptual simple harmonic motion question

1. Jul 3, 2010

### calgal260

1. The problem statement, all variables and given/known data
A block is attached to a horizontal spring. On top of this block rests another block. The two-block system slides back and forth in simple harmonic motion on a frictionless horizontal surface. At one extreme end of the oscillation cycle, where the blocks come to a momentary halt before reversing the direction on their motion, the top block is suddenly lifted vertically upward, without changing the zero velocity of the bottom block. The simple harmonic motion then continues. What happens to the amplitude and the angular frequency of the ensuing motion?

2. Relevant equations
PE(elastic)= 0.5kA^2

3. The attempt at a solution

The answer that they give us is amplitude remains the same, while angular frequency increases.

I understand the angular frequency part but not the amplitude part. When the spring sets in motion and you look at the total energy equation, shouldn't the elastic potential energy be less now, now that angular frequency is no longer zero (since it's not halted) and you have rotational and translational kinetic energy? And they tell you that there is no friction, which means energy is conserved. How can energy be conserved when the potential elastic energy (from which the amplitude is derived) remains the same, and thus amplitude as well?

Here is the book explanation. It still confuses me:
At the instant the top block is removed, the total mechanical energy of the remaining system is all elastic potential energy and is 12 kA2 (see Equation 10.13), where A is the amplitude of the previous simple harmonic motion. This total mechanical energy is conserved, because friction is absent. Therefore, the total mechanical energy of the ensuing simple harmonic motion is also 12 kA2 , and the amplitude remains the same as it was previously. The angular frequency ω is given by Equation 10.11 as ω = mk . Thus, when the mass m attached to the spring decreases, the angular frequency increases.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jul 3, 2010

### vela

Staff Emeritus
The angular frequency $\omega=\sqrt{k/m}$ is never zero, and there's no rotational kinetic energy in the problem. This is a one-dimensional problem, so there's no angular velocity either. You meant the linear velocity of the block is no longer zero, and yes, once the block starts moving again, the potential energy must decrease for the total mechanical energy to remain constant.
I'm not sure what you're getting at here, but I suspect it has to do with confusion over what amplitude means. What specifically do you mean when you are referring to the amplitude?

3. Jul 3, 2010

### calgal260

That is the maximum distance from which the spring may stretch to from equilibrium? Or can this also be referred to as the distance you pull it by?

So if the rotational kinetic energy is never zero, is its moment of inertia always constant?

4. Jul 3, 2010

### vela

Staff Emeritus
Yes, that's correct. So I'm not sure what you were getting at when you asked how energy could be conserved. Could you elaborate on where you think the problem is?
This problem has absolutely nothing to do with rotational motion. You just have a block connected to a spring sliding back and forth along a straight line on a frictionless table. You're probably getting confused because some of the mathematics for rotational motion and simple harmonic motion is similar.

5. Jul 3, 2010

### calgal260

Most of my confusion comes from trying to match my logic with that of my textbook's solution. Here's what they said: At the instant the top block is removed, the total mechanical energy of the remaining system is all elastic potential energy and is 12 kA2 (see Equation 10.13), where A is the amplitude of the previous simple harmonic motion. This total mechanical energy is conserved, because friction is absent. Therefore, the total mechanical energy of the ensuing simple harmonic motion is also 12 kA2 , and the amplitude remains the same as it was previously. The angular frequency ω is given by Equation 10.11 as ω = mk . Thus, when the mass m attached to the spring decreases, the angular frequency increases.

6. Jul 3, 2010

### vela

Staff Emeritus
Yeah, I saw the book said earlier. I just can't tell where you see a contradiction. You asked, "How can energy be conserved when the potential elastic energy (from which the amplitude is derived) remains the same, and thus amplitude as well?" My question to you is, how can energy not be conserved if the potential energy is unchanged when the block is removed? It was $1/2\,kA^2$ both before and after, so it's constant, i.e. conserved.