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.