Determine the equation of motion of a point mass m attached to a coiled spring with spring constant k, initially displaced a distance x, from equilibrium, and released with velocity v, subject to no damping or other external forces. 1. m= 10kg , k = 1000/m, x, = 1m., v, = 0

Explanation:

Angular velocity w =

m = 10 ;  k = 1000N  / m  

Putting these values we get

w = 0.1

Amplitude  A  = 1 m  

So equation of  S H M is

x = A cos ( wt )

Velocity dx / dt = Aw sin wt  when t = 0 , velocity = 0 (given )

The equation for motion is

x = 1 x cos ( 0.1 t )

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this