# Damped Oscillations & Sequences

• October 21st 2007, 08:39 PM
xfyz
Damped Oscillations & Sequences
1) Spring of mass 1kg and its damping constant c = 10. The spring starts from it equilibrium position with a velocity of 1 m/s. graph the position function for the following values of the spring constant k: 10,20,25,30,40. What type of damping occurs in each case?

so m = 1kg, c =10, x(0) = 0 and x'(0) = 1

1r^2 + 10r + kx = 0.
do i calculate each one by using the given k values? is it necessary to use 10 pages?

2)Can someone show me how to figure out the formula mathematically?
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• October 22nd 2007, 06:47 AM
topsquark
Quote:

Originally Posted by xfyz
1) Spring of mass 1kg and its damping constant c = 10. The spring starts from it equilibrium position with a velocity of 1 m/s. graph the position function for the following values of the spring constant k: 10,20,25,30,40. What type of damping occurs in each case?

so m = 1kg, c =10, x(0) = 0 and x'(0) = 1

1r^2 + 10r + kx = 0.
do i calculate each one by using the given k values? is it necessary to use 10 pages?

The equation for a damped spring is
$\frac{d^x}{dt^2} + \frac{c}{m} \cdot \frac{dx}{dt} + \frac{k}{m} \cdot x = 0$
Where the damping force is given as $F = -c \frac{dx}{dt}$.

This is a homogeneous linear differential equation so the auxiliary equation is
$r^2 + \frac{c}{m} \cdot r + \frac{k}{m} = 0$

The solution to this equation is (using the quadratic formula)
$r = \frac{ -\frac{c}{m} \pm \sqrt{ \frac{c^2}{m^2} - 4(1) \cdot \frac{k}{m}}}{2}$

$r = \frac{-c \pm \sqrt{c^2 - 4km}}{2m}$

So the general solution for the differential equation is
$x(t) = A \cdot exp \left [ x \left ( \frac{-c + \sqrt{c^2 - 4km}}{2m} \right ) \right ] + B \cdot exp \left [ x \left ( \frac{-c - \sqrt{c^2 - 4km}}{2m} \right ) \right ]$

So the nature of the discriminant will determine the nature of the solution.

Your problem specifies m = 1 kg and c = 10 Ns/m. (You didn't specify the units for this, so I'm presuming they are MKS units.)

So your solution comes out to be:
$x(t) = \left ( \frac{1}{2\sqrt{25 - k}} \right ) \cdot exp \left [ x \left ( \frac{-10 + \sqrt{100 - 4k}}{2m} \right ) \right ]$ $- \left ( \frac{1}{2\sqrt{25 - k}} \right ) \cdot exp \left [ x \left ( \frac{-10 - \sqrt{100 - 4k}}{2m} \right ) \right ]$

The nature of the solution changes depending on the sign of the discriminant. If the discriminant is negative we have damped harmonic motion. If the discriminant is 0 we have critically damped harmonic motion. If the discriminant if positive we have overdamped harmonic motion.

-Dn