# Harmonic Oscillator - Mass With Initial Velocity

1. Feb 22, 2012

### Eugbug

For a harmonic oscillator with mass M, spring of stiffness k and displacement the force equation is:

-kx = Md2x/dt2

How do you handle the situation and work out a solution for x(t) when the mass has an initial velocity. E.g. a mass dropped onto the spring?

2. Feb 22, 2012

### genericusrnme

It's a pretty simple second order differential equation with sins and coss as it's solutions, you then just solve for x'(0) = initial velocity

3. Feb 22, 2012

### Philip Wood

The first thing you do is to solve the DE in the general case, that is without specific initial conditions. You find
x = A cos $\omega$t + B sin $\omega$t

in which $\omega$ = $\frac{force per unit x}{mass}$

and A and B are arbitrary constants. It is these whose values accommodate the initial conditions. So, suppose you knew that at t = 0, x = 0 and $\frac{dx}{dt}$ = v$_{0}$.

Substituting x = 0, t = 0 into the general solution gives

x = B sin $\omega$t. [A = 0.]

So $\frac{dx}{dt}$ = B $\omega$ cos $\omega$t

Now imposing $\frac{dx}{dt}$ = v$_{0}$ when t = 0

v$_{0}$ = B $\omega$ so B =$\frac{v}{\omega}$

Putting in this value for B, we finally have

x = $\frac{v}{\omega}$ sin $\omega$t.

4. Feb 22, 2012

### Eugbug

Thanks for the answer Philip!

I was reading a question about Bungee ropes last night on Webanswers and this equation could probably go some way towards modeling the behavior of a jumper when they have reached the end of the rope which then starts to stretch.

5. Feb 22, 2012

### Philip Wood

Yes, it would be an excellent model. You've got to be careful over gravity, which displaces the equilibrium position below the point at which the bungee starts to stretch, but I doubt if this will cause too many problems for you.

And – not something to worry about yet – the bungee won't obey Hooke's law perfectly.