A Mass weighinh 5 ibs stretches a spring 3 inches, bringing it to equalibrium. T

Question

A Mass weighinh 5 ibs stretches a spring 3 inches, bringing it to equalibrium. The mass is then pushed upward from equilibrium, contracting the spring a distance of 2 inches, and then set in motion with a downward velosity of 2 ft/sec at t=0. Assume no damping is present.

A) Find the diplacement, X, of the mass as a function of time,t.

b) what is the maximum displacement of the mass from equilibrium?

c) How much time is required to complete one full oscillation of the mass-- the period of the displacement function?

Explanation / Answer

The equilibrium position of the motion is still at the stretch-point of 3 inches.

However, the initial conditions are:
x(0) = 2 inch
x'(0) = 2 ft/s

The differential equation is:
mx'' + kx = 0
x'' + (k/m) x= 0

So:
x(t) = A*sin(wt + P)
x'(t) = A*w*cos(wt + P)

where
w = sqrt(k/m)

The only real challenge with this problem is getting this into workable units.

mg = 5 lbs = 5*4.448 N
So:
m = 5*4.448/9.8 = 2.26 kg
k = 5 lb/3 in = (5*4.448)/(3*0.0254 m)
= 293.96 N/m
w = sqrt(k/m) = 11.40 rad/s

I'm gonna sketch it out and leave it for you to finish:

- You need to convert x(0) and x'(0) into metric units, so that the formulas will not be confused.

- Then you know x(0) and x'(0) in terms of A, w and P; and you know w.

So you can find A by squaring x(0) and adding (x'(0)/w), and then you can find P by taking the ratio of x'(0) and

x(0), and taking the arctangent.

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