Answer to Question #252247 in Mechanics | Relativity for Anuj

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

Solution.

$F=kx\implies k=\dfrac{F}{x}=\dfrac{2}{1}=2;$

$W=mg\implies 3.2=32m\implies m=0.1 slugs.$

$\beta=0.4;$

$mx^{''}+\beta x^{'}+kx=0;$

$0.1x^{''}+0.4x^{'}+2x=0;$

$x^{''}+4x^{'}+20x=0;$

$2\lambda=4; \lambda=2, \omega^2=20$

$x(t)=e^{-\lambda t}(C_1cos(\sqrt({\omega^2-\lambda^2}t)+C_2sin(\sqrt{\omega^2-\lambda^2}t);$

$x(t)=e^{2 t}(C_1cos(4t)+C_2sin(4t);$

$x^{'}(0)=0; x(0)=-1; C_1=-1;$

$x^{'}(0)=-2C_1+4C_2=2+4C_2\implies C_2=-\dfrac{1}{2};$

$x(t)=e^{-2t}(-cos(4t)-\dfrac{1}{2}sin(4t));$

$x(\dfrac{\pi}{2})=e^{-\pi}(-cos(2\pi)-\dfrac{1}{2}sin(2\pi))=e^{-\pi}.$

Answer: $a)x(t)=e^{-2t}(-cos(4t)-\dfrac{1}{2}sin(4t));$

$b)e^{-\pi}.$