Find its displacement?

Question

A 64 lb object stretches a spring 4 ft in equilibrium. It is attached to a dashpot with damping constant c=8 lb-sec/ft. The object is initially displaced 18 inches above equilibrium and given a downward velocity of 4 ft/sec. Find its displacement and time-varying amplitude for t>0.

 

The equation of motion is my"+cy'+ky=0. How do I find m, c and k? Please show all your work.

Answer 1

k = -F/x₀ = -(64 lb)(-4 ft) = 16 lb/ft

 

c = 4 lb·s/ft (given)

 

m = F/g = (64 lb)/(32.174 ft/s²) ≈ 2 lb·s²/ft

 

We can now solve the differential equation.

 

2y'' + 4y' + 16y = 0

 

y'' + 2y' + 8y = 0

 

λ² + 2λ + 8 = 0

 

λ = -1 ± i√7 = -1 ± 2.646i

 

y = e^-t[A sin(2.646t) + B cos(2.646t)]

 

We are given y(0) = 18 in = 1.5 ft and y'(0) = -4 ft/s.

 

y(0) = e⁰[A sin 0 + B cos 0] = B = 1.5

 

y' = -2.646e^-t[A sin(2.646t) + B cos(2.646t)] + 2.646e^-t[A cos(2.646t) - B sin(2.646t)]

 

y'(0) = -2.646e⁰[A sin 0 + B cos 0] + 2.646e⁰[A cos 0 - B sin 0] = 2.646(A - B) = -4

 

A - B = -4/2.646 = -1.512

 

A = -1.512 + B = -0.012 ≈ 0

 

Plugging in A ≈ 0 ft. and B = 1.5 ft. gives

 

y = 1.5e^-tcos(2.646t)