A hanging spring is stretched 6 inches 5 feet by a weight of

A hanging spring is stretched 6 inches (=.5 feet) by a weight of 64 pounds. Set up the initial value problem (differential equation and initial conditions\' which describes the motion, neglecting friction, when the weight is pulled down an additional foot and is then released. You need not solve the equation. (Recall that g, the acceleration due to gravity is 32 feet/second2.)

Solution

First we compute the spring constant using data given in first line

Let spring constant be : k pound per foot

So,

64=k*0.5

Hence, k=128 pound /foot

This inital stretch by 0.5 feet is the equilibrium position of the spring where spring force and mass of weight cancel each other.

The equation of motion is then set up for displacement about this equilibrium position, x(t)

So initial condition is:

x(0)= 1 foot

x\'(0)= 0   (This means initial speed is zero)

(We take displacement of pulling down from equilibrium as positive and moving above equilibrium as negative)

As weight is displaced x unit down. There is a restoring force: kx acting up and mg acting down

Denoting derivative w.r.t. time by \'

The acceleration becomes: x\'\'

x is positive in downward direction so acceleration is positive downwards.

So,

mx\'\'=60-kx

Or,

x\'\'=32-32*2x=32(1-2x)

x\'\'=32(1-2x),

x(0)= 1 ,x\'(0)= 0