Check out some similar questions!

ilovemath · Apr 20, 2012, 05:50 PM

A force of 64 N stretches a spring 4 meters. The mass of 4 kg is attached to the spring and set into motion in a medium that offer a damped force equal 16 times the velocity .If the mass is pulled down 2 centimeters from the equilibrium position and released. Find the equation for the position. Find the time(s) at which the mass passes the equilibrium position.

ebaines · Apr 20, 2012, 06:13 PM

The basic differenjtial equation for a mass, spring and damper is

$<br /> m \frac {d^2x}{dt^2} +C\frac {dx}{dt} + kx = 0<br />$

You know that k = 64N/4m = 16t N/m and C = 16N/(m/s). Set $x = Ae^{bt}$ and solve for A and b. You should end up with a solution for b using the quadratic equation along the lines of
$<br /> b = \frac {-c} {2m} \pm \sqrt {\frac {c^2}{4m^2} - \frac k m}<br />$

Recall that if b is a complex number then the angular velocity $\omega$ is the imaginary part of the value for b and the damping constant is the real part. The value for A comes from the initial condition that x(0) = 0.02m. If you need more help post back with your work and we can help you along.