# Math Help - Solution for t of displacement of a mass, spring, damper system

1. ## Solution for t of displacement of a mass, spring, damper system

Hi, i'm trying to re-arrange the standard solution for displacement of an oscillating mass spring damper system so that it gives time (t) in terms of displacement (y).

The solution I currently have for y is:

y = e^(At).(C.sin(Bt)+D.cos(Bt))

Can anyone give me a solution for t????

(e = Eulers number; A,B,C & D are constants)

Thanks,

Dan

2. A couple of thoughts come to mind. First: what is your original DE? Do you know what the spring constant is? Do you know what the damping force is exactly?

Second, what are the initial conditions?

3. Originally Posted by DannyH
y = e^(At).(C.sin(Bt)+D.cos(Bt))
This cannot be analytically solved for t. The equation with which I am more familiar with:
$\displaystyle mx^{\prime \prime} + b x^{\prime} + kx = 0$

has, for a standard solution:
$\displaystyle x(t) = A e^{-bt/(2m)}~cos( \omega t + \delta)$

cannot, in general, be solved for t either.

You're stuck with numerical methods for this one, I'm afraid.

-Dan

Come to think of it, my form and yours are essentially the same. All you need to do is expand the cosine in mine and you get your form.

4. Originally Posted by topsquark
This cannot be analytically solved for t. The equation with which I am more familiar with:
$\displaystyle mx^{\prime \prime} + b x^{\prime} + kx = 0$

has, for a standard solution:
$\displaystyle x(t) = A e^{-bt/(2m)}~cos( \omega t + \delta)$

cannot, in general, be solved for t either.

You're stuck with numerical methods for this one, I'm afraid.

-Dan

Come to think of it, my form and yours are essentially the same. All you need to do is expand the cosine in mine and you get your form.
Do you know it cannot be solved for analytically by just looking at it or is it based on experience?

5. Originally Posted by bugatti79
Do you know it cannot be solved for analytically by just looking at it or is it based on experience?
In this case, experience. However it doesn't take long to have some knowledge in this area. Any equation that involves something like x + f(x) = 0 where f(x) is a trig function, exponential, or logarithm can immediately probably does not have an analytical solution. (Though some of these do have "accidental" solutions...solutions that just happen to be integers. But these can only be found by inspection. For example 3^(x-1) - x = 0 has two solutions, the first is x = 1 but the other has to be approximated.)

-Dan

6. Thnaks for this guys, saved me lots of watsed time trying to figure that out.

However i think i solved it for the case where y = 0 (which fortunately is a key point in what i'm trying to find). Unfortunately the result doesn't match up with my numerical results. If you could take a look below and let me know where of gone wrong it would be great....

y = e^(At).(C.sin(Bt)+D.cos(Bt))

Lets say E is the angle who's cosine is D/SQRT(C^2+D^2)

so D = cos(E).SQRT(C^2+D^2)
C = sin(E).SQRT(C^2+D^2)

substitute these into the original equation to get

y = e^(At).( sin(E).sin(Bt).SQRT(C^2+D^2) + cos(E).cos(Bt).SQRT(C^2+D^2) )

= e^(At).( [cos(E-Bt)-cos(E+Bt)]/2 + [cos(E+Bt)+cos(E-Bt)]/2 ).SQRT(C^2+D^2)

= e^(At).cos(E-Bt).SQRT(C^2+D^2)

now if y = 0 then either

e^(At)= 0 - no solution

SQRT(C^2+D^2) = 0 - only has the imaginary solution D = iC (i think this is right but it's something close and is not the case, C and D are always real)

so I'm left with:

cos(E-Bt) = 0

therefore for y= 0

t = (E-arccos(0))/B

= (arccos(D/SQRT(D^2+C^2)) - arccos(0))/B

As i mentioned this doesn't tally with my numerical result for any given values of A,B,C and D so any help would be really appreciated.

Dan

7. Originally Posted by DannyH
y = e^(At).(C.sin(Bt)+D.cos(Bt))
now if y = 0 then either
Yes this can be solved if y = 0, but this is a very special case. It cannot be solved for arbitrary values of y.

-Dan