# Mass on a horizontal spring

1. Jun 19, 2009

### sarah895

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

A 50 gram mass is attached to a massless spring and allowed to oscillate around an equilibrium according to:
y(t) = 1.2*sin( 3.1415*t ) where y is measured in meters and t in seconds

(a) What is the spring constant in N/m ?
(b) What is the total Mechanical Energy in the mass/spring system?
(b) What is the maximum kinetic energy of the mass?

2. Relevant equations

Fcirc=m*A*ω2.
F(x)=k*x
Fcirc=F(A)

3. The attempt at a solution

In an attempt to solve for the spring constant, I solved for the circular force (maw^2 = .592) and set that equal to k*x. For the x variable, I used the given equation y(t) = 1.2*sin( 3.1415*t ). Since we don't know t, I assumed it to be 1. I then solved F=kx for k, and got an answer of 9N/m, which was wrong.

Any guidance would be much appreciated :)

2. Jun 19, 2009

### cepheid

Staff Emeritus
EDIT: Welcome to PF!

I'm sorry, but I can't seem to make much sense of your approach. Nothing is moving in a circle here, so what is the circular force supposed to be? Why have you listed it as relevant?

Oscillatory motion in general is described by a function of the form

y(t) = Asin(ωt)​

In this case, you have been given values for A and ω. So you know ω. What is the relationship between ω and k for a spring mass system?

3. Jun 19, 2009

### cepheid

Staff Emeritus
I just thought I'd explain why this is not correct. In this case, t is the independent variable. Emphasis on the word variable. It changes, it can be whatever you want it to be. The position, y, is the dependent variable, because its value depends on the value of t based on the mathematical relationship given. Intuitively this makes sense. The position of the mass depends upon time because the mass is oscillating. We say that y is a function of t. What this means is that you can plug in any value of t into y(t), and you will get a corresponding value of y that represents the position of the mass at that time. Do you understand?