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.) Find the first time at which the mass passes through the equilibrium position
heading upwards.

t =1.294

Explanation:

Let m be the mass attached, let k be the spring constant and let p be the positive damping constant. The Newton's second law for the system is

where x(t) is the displacement from the equilibrium position. The equation can be transformed into  

(a) Let's determine the equation of motion. We must convert units of weight into units of mass  

From Hooke's law we can calculate the spring constant k.  

If we put m = 1/10 slugs, k = 2 lb/ft and = 0.4 into the DE, we get

The auxiliary equation is m^2+4*m+20 = 0 and its solutions are m_1 = -2-4_i and m_2 = -2 + 4i. The general solution is then  

From the initial conditions

we can find the equation of motion  

(b) We need to express the equation of motion in the alternative form

first attachment

The amplitude is  

and the phase angle is

sin∅=

cos∅=

tan∅=2

∅=

 =4.24

(c) lb find the wanted time, we will need the derivative of the equation of motion  

x'(t)=

The mass passes through the equilibrium when  

x(t) =0

t =-0.2786+k/4

The velocity at those times is  

second attachment

The first (positive) time at which the mass passes through the equilibrium heading upward is for k = 2 (when x' < 0).  

t =1.294

0.19 m/s and 0.45 m/s

pythagoras' theorem

square root (-5^2+9^2)

square root (25+81)

square root (106)

angle is tan^-1(9/5)

neg x, pos y