Question

 Consider a 1kg mass suspended to a spring. A force of 20 Newtons is required to stretch it 5 meters below it’s natural length. (a) Determine the equation describing the motion of the system, if the system is released 1 meter below it’s natural position, with a push of 5 m/s upward, and damping is such that b = 2√ 8. 

Answer 1

$mx''+bx'+kx=0$

 characteristic equation:

$m\lambda^2+b\lambda +k=0$

$\lambda=\frac{-b\pm \sqrt{b^2-4km}}{2m}$

$k=20\ N/5\ m=4$ N/m

$\lambda=\frac{-2\sqrt 8\pm \sqrt{32-16}}{2m}$

$\lambda=-\sqrt 8\pm 2$

$x(t)=c_1e^{\lambda_1t}+c_2e^{\lambda_2t}=c_1e^{(-\sqrt 8- 2)t}+c_2e^{(-\sqrt 8+ 2)t}$

$x(0)=c_1+c_2=1$

$v(t)=x'(t)=(-\sqrt 8- 2)c_1e^{(-\sqrt 8- 2)t}+(-\sqrt 8+ 2)c_2e^{(-\sqrt 8+ 2)t}$

$v(0)=(-\sqrt 8- 2)c_1+(-\sqrt 8+ 2)c_2=-5$

$(-\sqrt 8- 2)(1-c_2)+(-\sqrt 8+ 2)c_2=-5$

$-\sqrt 8-2+4c_2=-5$

$c_2=(\sqrt 8-3)/4,c_1=1-(\sqrt 8-3)/4=(7-\sqrt 8)/4$

$x(t)=\frac{7-\sqrt 8}{4}e^{(-\sqrt 8- 2)t}+\frac{\sqrt 8-3}{4}e^{(-\sqrt 8+ 2)t}$