# Scaling Equations

1. Dec 11, 2008

### ACE_99

1. The problem statement, all variables and given/known data
Many undamped mechanical vibrations are described by the Differential Equation

x′′ + Kx = 0

that describes a position x(t) with t being time and K a positive constant.

a) What are the MKS units for K?
b) Introduce s=t/T for some characteristic time T to be determined (i.e. scale time) to make the equation for x(s) as simple as possible.

3. The attempt at a solution
Im not sure where to start with this problem. Any help would be greatly appreciated. Im also having some problems understanding scaling (non-dimentionalization) so If some one could help with that to it would be great.

2. Dec 11, 2008

### HallsofIvy

Staff Emeritus
Yes, I can see that if you cannot do this problem, you must be completely stumped by scaling! Think about the other parts of the equation. "x" is a distance so is measured (in MKS units) in meters. "t" is a time so is measured in seconds. The derivative is a limit of the "difference quotient", (x(t+h)- x(t))/h and since h is added to t, is a time, measured in seconds itself. That derivative has units of meters/seconds. Which makes sense because the derivative of distance with respect to time is "speed". You can argue the same way again: the second derivative is the derivative of the first which involves the difference quotient (x'(t+h)- x'(t))/h, (m/s)/s= m/s2. Or you can just think "second derivative of distance with respect to time is acceleration so it has units of m/s2.

That means your equation, in terms of its units is
x" (m/s2)+ k (?) x (m)= 0.
Of course, to be able to add those and get 0, the two parts, x" and kx, must have the same units. m/s2= ?*m. What must ? be to make that true?

3. Dec 11, 2008

### ACE_99

So ? should be 1/s2. Making the units of k 1/s2

4. Dec 11, 2008

### HallsofIvy

Staff Emeritus
Yes. Notice also that that is almost the "spring equation". Multiplying by m, mx"+ mkx= 0 or mx"= -mk x which says "mass times acceleration" = "force"= -"spring constant times distance stretched. The spring constant, of course is given by "Newtons per meter". Since a Newton is 1 kg m/s2, "Newtons per meter" is just kg/s2 and taking out the kg again, leaves k in terms of 1/s2.

5. Dec 11, 2008

### ACE_99

Great that helps me with one part of the question. Any tips on the time scale part?

6. Dec 12, 2008

### HallsofIvy

Staff Emeritus
About the only way you can "simplify" this is to get rid of that constant, K. If you take s= t/T, then dx/dt= dx/ds (ds/dt)= (1/T)dx/ds. d2x/dt2= d(x')dt= (1/T2)d2x/ds2. Put that into the equation and see what T must be in order to cancel K.

Since the solutions to this equation (for K positive) will be periodic, an obvious choice of time interval would be the period. Is the T you got above related to the period of the solutions?