Question

A force of 2 pounds stretches a spring 1 foot. A mass weighing 3.2 pounds is attached to the

spring, and the system is then immersed in a medium that offers a damping force that is

numerically equal to 0.4 times the instantaneous velocity. (a) Find the equation of motion

if the mass is initially released from rest from a point 1 foot above the equilibrium position.

(b) Determine the position and velocity of the mass at t = pi/2 sec? (c) Find the first time at

which the mass passes through the equilibrium position heading upward.

Answer 1

$F_1= 2lbf; x_1=1ft$

$k = \frac{F_1}{x_1}=2\frac{lbf}{ft}$

$F_g= 3.2lbf;g\approx32\frac{ft}{s^2}$

$m = \frac{F_g}{g}=0.1lb$

$a)$

$F= ma$

$a(t) = x''(t)$

$v(t) =x'(t)$

$x(0)= -1ft;v(0)=x'(0)= 0$

$F_{spr}= kx$

$F_{dmp}= 0.4*v(t)= 0.4x'(t)$

$F+F_{spr}+F_{dmp}=0$

$ma +kx+0.4x'(t)=0$

$0.1x''(t)+0.4x'(t) +2x = 0$

$x''(t)+4x'(t)+20=0$

$x = e^{rt}$

$r^2+4r+20=0$

$r_1= -2+4*i;r_2 = -2-4*i$

$x_1(t)= e^{-2t}\cos(4t)$

$x_2(t)= e^{-2t}\sin(4t)$

$x(t) = C_1 e^{-2t}\cos(4t)+C_2 e^{-2t}\sin(4t)$

$x(0) = -1:C_1 =-1$

$x'(t) = -4C_1e^{-2t}\sin(4t)-2C_1e^{-2t}\cos(4t)\newline -C_2e^{-2t}\sin(4t)-4C_2e^{-2t}\cos(4t)$

$x'(0)=0$

$-2C_1+4C_2= 0$

$C_2= \frac{C_1}{2}=-0.5$

$x(t) = - e^{-2t}\cos(4t)-0.5 e^{-2t}\sin(4t)$

$b)$

$t = \frac{\pi}{2}s$

$x(t) = - e^{-2t}\cos(4t)-0.5 e^{-2t}\sin(4t)$

$x(\frac{\pi}{2})=- e^{-\pi}=-0,043ft$

$x'(t) = 4e^{-2t}\sin(4t)+2e^{-2t}\cos(4t)\newline +0.5e^{-2t}\sin(4t)+2e^{-2t}\cos(4t)$

$v(\frac{\pi}{2})=4*e^{-\pi}=0.173\frac{ft}{s}$

$c)$

$x(t)= 0$

$x(t) = - e^{-2t}\cos(4t)-0.5 e^{-2t}\sin(4t)$

$- e^{-2t}\cos(4t)-0.5 e^{-2t}\sin(4t)= 0$

$cos(4t)+0.5sin(4t)= 0$

$\tan(4t) = -2$

$t = \frac{\pi n}{4}-\frac{\arctan2}{4};t>0$

$\text{the mass passes through the equilibrium}$

$\text{position in the upward direction}$

$\text{by condition, 2 value is required } t\text{ where } t>0$

$t = \frac{2\pi }{4}-\frac{\arctan2}{4}\approx1.3s$