Question

A 1 newton force will stretch a spring 1 meter. The spring/mass system is damped by a force that is 8 times the instantaneous velocity. A 16 kg mass is attached to the spring. The spring is compressed 0.6 meters above the equilibrium position and given an initial downward velocity of 2 m/s. Determine the equation of motion of the mass

Answer 1

Answer:

Step-by-step explanation:

Given the mass is m =16kg, and 1N force will stretch the spring 1 m.

That is, F =1N,Z =1m. Now find the spring constant k:

F = kL = 1 = k(1) = k= 1N/m.

The damping force is 8times the instantaneous velocity, this means β = 8,

and the external force is f(t) = 0

Initially the object compressed 0.6m above equilibrium position,

with the downward velocity is 2m/s.

The differential equation for a spring mass system with

damping force and extemal force is: mx" + βxt + kx = f(t).

so, 16x"+ 8x' + x= 0, x(0} = -0.6, x'(0)= 2m/s.

Now solve the DE:

The auxilary equation for the homogeneous equation is 16x"+8x'+x=0

solving we get, 16r² + 8r + 1 = 0 => (4r + 1)² = 0 => r = - 1/4.

Then the general solution for the homogenous system is: $x(t)=c_1e^{-t/4} +c_2te^{-t\4}$.

Use the initial conditions x (0) = -0.6, x'(0) = 2m/s:

$x(0)=c_1e^{0} +c_2(0)e^{0}=-0.6=c_1\\x'(t)=-\frac{1}{4}c_1e^{-t/4}+c_2e^{-t/4}-\frac{1}{4}c_2te^{-t/4}\\x'(0)=-\frac{1}{4}c_1e^0+c_2e^0-\frac{1}{4}c_2(0)e^0=2=-\frac{1}{4}(-0.6)+c_2=c_2=1.85$.

Hence, $x(t) =-0.6e^{-t/4}+1.85te^{-t/4}$.