there is a weight attached to the end of a spring its lifted

there is a weight attached to the end of a spring it\'s lifted 6cm above its resting point and dropped and takes .5 seconds to return to its starting position. write an equation for the path of weight

Solution

first

mg -kx = ma

a= g - (k/m)x

but a(accelaration) = dx^2 /dt^2

so  dx^2 /dt^2 = g - (k/m)x

now integrate with respect to t

dx/dt = gt - (k/m)xt

integrate again

x = gt^2 / 2 - (k/m)xt^2 / 2

here x= 6cm and t =0.5 , and take g=10

now find k/m value

0.06 = 10(0.5)^2 /2 - (k/m) 0.06 (0.5)^2 /2

0.06 = 5*0.25 - (k/m) 0.03(0.25)

0.06 = 1.25 - (k/m) 0.75

-(k/m) 0.75 = 0.06 - 1.25

-(k/m) 0.75 = -1.19

(k/m) = 1.61

now plug this value in the first equation

a = g - (1.61) x