A force of 400 newtons stretches a spring 2 meters A mass of

A force of 400 newtons stretches a spring 2 meters. A mass of 50 kilograms is attached to the end of the spring and is initially released from the equilibrium position with an upward velocity of 6 m/s. Find the equation of motion. x(t) =

Solution

Ideal Springs and masses create simple harmonic motion.
Which runs on an equation like
y(t) = Asin(t)

We need to find A and

The first statement tell us the spring constant

k = 400 N / 2 m = 200 N/m

The kinetic energy at the neutral position will be converted to Potential energy at the highest vertical point. Ignore friction

PE = KE

mgh = ½kx²

but if we set our origin for height at the neutral point then h = x

mgx = ½kx²

mg = ½kx

x = 2mg/k

x = 2(50)(9.8)/200

x = 4.9 m

which will also be the maximum amplitude A for our equation

y(t) = 4.9sin(t)

from SHM studies we also know that

= (k/m)

= (200 / 50)

= 2 rad/s

our equation for motion is

y(t) = 4.9sin(2t)

y(/12) = 4.9sin(2(/12))

y(/12) = 2.5 m