Answer to Question #138205 in Differential Equations for Nikhil Singh

Question

A vibrational system consisting of mass 1/10 kg is attached to a spring (spring constant= 4kg/m). The mass is released from rest 1m below the equilibrium position. the motion is damped (damping constant=1.2) and is being driven by an external force 5sin4t, beginning at t=0. Write the governing equations of the system and intrepet the type of motion. Hence find the position of mass at time t.

Answer 1

Given,

Mass m=0.1kg

spring constant k=4kg/m

displacement of mass s=1m

Damping constant c=1.2

External force F=5sin4t

Damping force , $F_d=-cs'$ (where s' is velocity)

Spring force f$=-k(x+s)$

Force due to gravity on the block= mg

When the Mass is in equillibrium

According to hook's law

mg=ks

Now the equation that governs the system is given by,

$mx''=mg-k(s+x)+f_d+F$

$mx''=-kx-cx'+5sin4t$ ( since mg=ks)

m$x''+cx'+kx=5sin4t$ ....(1)

Now Damping ratio is given by,

$\zeta=\frac{c}{2\sqrt{mk}}$

=$\frac{1.2}{2\times \sqrt{0.1\times 4}}$

=$\frac{0.6}{\sqrt{0.4}}$

=0.94<1

$\zeta<1$

So The motion is underdamped.

for position solving equation equation 1,

0.1x''+1.2x'+4x=5sin4t

on solving the left side part,

$0.1m^2+1.2m+4=0$

Roots of above equation is given by,

$r_1=-\frac{c}{2m}-i\omega_1$ and $r_2=-\frac{c}{2m}+i\omega_1$

where $\omega_1=\frac{\sqrt{4mk-c^2}}{2m}$ =$\frac{\sqrt{4\times 0.1\times 4-(1.2)^2}}{2\times0.1}$

=$\frac{1.6-1.44}{0.2}$ =$\frac{0.16}{0.2}$ =0.8

The complimentary function is given by,

c.f=$e^{-\frac{ct}{2m}}(c_1cos0.8t+c_2sin0.8t)$

= $e^{-\frac{1.2t}{0.2}}(c_1cos0.8t+c_2sin0.8t)$

Particular integral is given by,

P.I.=$5\frac{sin4t}{0.1m^2+1.2m+4}$

=$\frac{5sin4t}{-0.1\times16+1.2m+4}$

=$\frac{5sin4t}{1.2m+2.4}$

=$\frac{5sin4t}{1.2m+2.4}\times \frac{1.2m-2.4}{1.2m-2.4}$

=$\frac{(1.2m-2.4)5sin4t}{1.44m^2-5.76}$

=$\frac{(1.2m-2.4)5sin4t}{1.44\times -16-5.76}$

=$\frac{(24cos4t-12sin4t)}{-23.04-5.76}$

=$\frac{(24cos4t-12sin4t)}{-28.8}$

Complete solution is given by,

x= $e^{-6t}(c_1cos0.8t+c_2sin0.8t)$ -$\frac{(24cos4t-12sin4t)}{28.8}$

This is the position at time t.