Understanding the Relationship Between Time and Position in a Mass-Spring System

In summary, the given equation describes the oscillatory motion of a 50 gram mass attached to a massless spring with a spring constant of 9N/m. The total mechanical energy in the system can be determined using the equation E = (1/2)kA^2, while the maximum kinetic energy of the mass can be found by substituting the maximum amplitude, A, into the equation for kinetic energy, KE = (1/2)mv^2. The value of t is not needed in these calculations, as the equations only depend on the given values of A and ω.
  • #1
sarah895
3
0

Homework Statement



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?


Homework Equations



Fcirc=m*A*ω2.
F(x)=k*x
Fcirc=F(A)

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 :)
 
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  • #2
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
sarah895 said:
Since we don't know t, I assumed it to be 1.

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?
 

FAQ: Understanding the Relationship Between Time and Position in a Mass-Spring System

What is the equation for the period of a mass on a horizontal spring?

The equation for the period of a mass on a horizontal spring is T = 2π√(m/k), where T is the period in seconds, m is the mass in kilograms, and k is the spring constant in Newtons per meter.

How does the mass affect the period of a horizontal spring?

The mass on a horizontal spring has a direct relationship with the period. As the mass increases, the period also increases. This means that a heavier mass will take longer to complete one oscillation on the spring.

Can the mass on a horizontal spring ever reach equilibrium?

Yes, the mass on a horizontal spring will reach equilibrium when the restoring force of the spring is equal to the weight of the mass. At this point, the mass will stop oscillating and remain at rest.

How does the spring constant affect the period of a horizontal spring?

The spring constant has an inverse relationship with the period. As the spring constant increases, the period decreases. This means that a stiffer spring will cause the mass to oscillate faster.

Is the motion of a mass on a horizontal spring considered simple harmonic motion?

Yes, the motion of a mass on a horizontal spring can be described as simple harmonic motion because it follows the principles of a restoring force that is directly proportional to the displacement from equilibrium.

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