# Homework Help: Simple Harmonic Motion: Giancoli Problem Help

1. Jan 10, 2014

### hstone

1. The problem statement, all variables and given/known data

A 1.15-kg mass oscillates according to the equation x = .650cos(8.40t) where x is in meters and t in seconds. Determine a)the amplitude, b)the frequency, c) the total energy of the system, and d) the kinetic and potential energy when x = 0.360m.

2. Relevant equations

x = Acosωt
ω = 2πf
Etotal = KE + PEs +PEg = 1/2mv² + 1/2kx² + mgy
v = rω
∑F = ma
F = kx

3. The attempt at a solution

a) Amplitude is given in the problem. A = .650m
b) Frequency f = ω/2π = 8.40/2π = 13.2 Hz
c) Total Energy E = 1/2mv² + 1/2kx² + mgy

This is where I am having difficulty, and feel like I am taking a complicated route and overlooking a simple solution. I assume there is no gravitational potential energy, since the problem doesn't mention any change in height. Simplifying and manipulating the equation for total energy I get the following:

E = 1/2m(Aω)² + 1/2(mg/x)(x²) = 1/2m[(Aω)² + gx]
E = 1/2(1.15)[(8.40*0.650)² + 9.8*0.360] = 19.2 J

I don't think this is correct, since I'm pretty sure I'm not supposed to use x = 0.360 in the equation. If I use the equation given in the problem I'd need to know t. I don't think this is the right way of going about this problem. I know I am not thinking about this correctly, and all of the possible equations and different forms are muddling my thought process.

Can I just calculate the total energy at ANY position to make this simpler? If so, I would use the equilibrium position x = 0, and my equation would be E = 1/2m(Aω)² = 17.14 J.

2. Jan 10, 2014

### vela

Staff Emeritus
Did you mean 1.32 Hz? Since $2\pi$ is greater than 1, numerically $f$ should be less than 8.40.

That's right. The only potential energy in this problem is the spring's.

Yes, you can. Energy is conserved, so if you calculate the total energy at any one place in the cycle, you know the total energy throughout the cycle.

One mistake you are making is using the relation $k = mg/x$. This applies to a mass hanging from a spring, where x is the amount the spring is stretched when the mass is in equilibrium. Since you don't have a mass hanging from a spring, you can't use that. You do have a relation that always applies to a mass-and-spring harmonic oscillator. You can use it to solve for $k$.

3. Jan 11, 2014

### hstone

When the spring force is the only force acting on the mass doesn't k = mg/x apply, since ΣF=kx=ma and the only acceleration is due to gravity? Should I be using ω=√(k/m), instead?

There are so many relations, I'm having difficulty keeping them all straight. I think I probably need to derive them all to be sure of where they are coming from.

Thanks so much for your help!! It's greatly appreciated.

4. Jan 11, 2014

### vela

Staff Emeritus
What are the forces on the mass? I'm assuming the problem is referring to a spring-mass system that moves horizontally. Is that right?

That'll work.

Deriving equations is good. It definitely helps you know when you can or can't use a particular equation.

5. Jan 11, 2014

### hstone

The problem just states that 'a mass oscillates.' It's pretty vague. I can see now why k=mg/x wouldn't work, since k is constant and shouldn't change with x.

Thanks!

6. Jan 11, 2014

### vela

Staff Emeritus
$k=mg/x$ applies to a mass hanging from a weight, and it only holds as the position where the force exerted by the spring, kx, equals in magnitude the weight of the mass, mg. In other words, it holds for a specific value of $x$.

I think it's safe to assume that you just have the archetypical harmonic oscillator, a mass connected to a spring and moving on a frictionless horizontal surface. Since the motion is horizontal, gravity doesn't play a role, which is another reason mg=kx wouldn't apply.