Harmonic Oscillator - Mass With Initial Velocity

[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?

 

[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

 

[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 \omegat + B sin \omegat

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 \omegat. [A = 0.]

So \frac{dx}{dt} = B \omega cos \omegat

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 \omegat.

 

[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.

 

[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.