Find its displacement?

Question

A weight stretches a spring 1.5 inches in equilibrium. The weight is initially displaced 8 inches above equilibrium and given a downward velocity of 4 ft/s. Find its displacement for t>0.

 

y"+(k/m)y=0

I need to convert the units to feet.

Answer 1

Use Hooke's Law: F = kx

 

The displacement x₀ of the equilibrium of the spring is given as x = 1.5 in = 0.125 ft. down.

 

Thus

 

F = kx₀ ⇒ m = kx₀/g ⇒ k/m = g/x₀ = (32.174 ft/s²)/(0.125 ft) = 257.392 s^-2

 

Solve the differential equation:

 

y'' = (k/m)y

 

y = A sin(ωt + φ)

 

where ω = √(k/m) is the angular frequency and φ is the phase.

 

We use the initial conditions: y(0) = 8 in = 0.667 ft, and y'(0) = -4 ft/s

 

A sin φ = 0.667 ft. ; Aω cos φ = -4 ft/s.

 

Take the ratio:

 

ω^-1 tan φ = -0.167 s.

 

tan φ = (-0.167 s)√(257.392 s^-2) = -2.67

 

φ = -1.21

 

sin φ = -0.94

 

A = (0.667 ft)/sin φ = (0.667 ft)/(-0.94) = -0.71 ft = -8.54 in.

 

In inches, the displacement is:

 

y(t) = -8.54 sin(16.04 t - 1.21)