Question

A 50g mass stretches the spring a distance of 20 cm.

1) Calculate the spring constant

2) Find the lowest damping constant r such that the trajectory of the mass has no oscillations

Answer 1

for 1) i found the spring constant to be 24.5 by doing:

mg = -k(x)

.49 = k*.2

k = 24.5

Answer 2

i'm a little lost with how to find the lowest damping constant. I have the equation:\[mx'' + rx' + kx = 0\]

Answer 3

so using the quadratic formula we can get \[\lambda = \frac{ -r \pm \sqrt{r^{2} - 4mk }}{ 2m } \]

and i remember something about how there needs to be repeated roots in order for there to be no oscilations... so do i just set r^2 -4mk to 0 and solve for r?

Answer 4

Well, consider whether our roots are *real* or *complex*. Complex roots correspond to complex exponentials in our transient solution, and if you're familiar with Euler's formula \(e^{ix}=\cos x+i\sin x\) corresponds to *oscillatory* functions :-) for *real* roots, we just have normal real exponentials in our solution which are monotone and do not oscillate

Answer 5

From our quadratic formula, the key is to check out what's under our radical, \(r^2-4mk\) -- also known as our discriminant \(\Delta\). If \(\Delta<0\) we end up with two complex roots and thus our solution will exhibit oscillatory behavior; if \(\Delta\ge0\) we end up with one or two *real* roots and thus solutions that exhibit monotonic behavior, i.e. something to the effect of: