Undamped Harmonic Motion (ODE problem)

[leroyjenkens]

Homework Statement

A 24-lb weight, attached to the end of a spring, stretches it 4 inches. Find the equation of motion if the weight is released from rest from a point 3 inches above the equilibrium position.

Homework Equations

\frac{d^{2}{x}}{dt^2}+\frac{k}{m}x=0
F=ma

The Attempt at a Solution

First I found the spring constant k using F=ma=ks at the point of equilibrium where the weight isn't moving. s is the displacement the weight causes on the spring, which is 4 inches, or 1/3 feet. So that equation turns out to be the weight equals the mass times the acceleration of gravity, which equals the spring constant times the displacement the weight caused, which is 24=k(\frac{1}{3}).

I found mass, m, by using F=ma which is 24lb=m(32\frac{ft}{s^2}) which gives me \frac{3}{4} slug

Putting that into my relevant equation at the top, it gives me \frac{d^{2}x}{dt^2}+96x=0 And using that one method (I forgot the name), I get m^2+96=0 which gives me m=\pm\sqrt{96}i

I use that for C_{1}cos\sqrt{96}t+c_{2}sin\sqrt{96}t

Now I have to find initial conditions to solve for C_{1} and C_{2}

My initial position is \frac{1}{4}ft above the equilibrium point, so does that mean I set that equation equal to -\frac{1}{4} and then solve for when t = 0?

And to get the other one, do I need to find the equation for the velocity by deriving the position equation? I'm not sure what I would set that equation equal to.

Thanks.

 

[Dick]

Homework Statement

A 24-lb weight, attached to the end of a spring, stretches it 4 inches. Find the equation of motion if the weight is released from rest from a point 3 inches above the equilibrium position.

Homework Equations

\frac{d^{2}{x}}{dt^2}+\frac{k}{m}x=0
F=ma

The Attempt at a Solution

First I found the spring constant k using F=ma=ks at the point of equilibrium where the weight isn't moving. s is the displacement the weight causes on the spring, which is 4 inches, or 1/3 feet. So that equation turns out to be the weight equals the mass times the acceleration of gravity, which equals the spring constant times the displacement the weight caused, which is 24=k(\frac{1}{3}).

I found mass, m, by using F=ma which is 24lb=m(32\frac{ft}{s^2}) which gives me \frac{3}{4} slug

Putting that into my relevant equation at the top, it gives me \frac{d^{2}x}{dt^2}+96x=0 And using that one method (I forgot the name), I get m^2+96=0 which gives me m=\pm\sqrt{96}i

I use that for C_{1}cos\sqrt{96}t+c_{2}sin\sqrt{96}t

Now I have to find initial conditions to solve for C_{1} and C_{2}

My initial position is \frac{1}{4}ft above the equilibrium point, so does that mean I set that equation equal to -\frac{1}{4} and then solve for when t = 0?

And to get the other one, do I need to find the equation for the velocity by deriving the position equation? I'm not sure what I would set that equation equal to.

Thanks.

Yes, if x(t) is the position then you know x(0)=(-1/4) (if you want displacements above equilibrium to be negative). Since it's release "from rest" you also know velocity at t=0 is 0. So set x'(0)=0.