Undamped Motion

Question

A weight stretches a spring 6 inches in equilibrium. The weight is initially displaced 4 inches above equilibrium and given a downward velocity of 3 ft/s. Find its displacement (in ft) for t>0 in terms of y(t)

Answer 1

I apologize. Sometimes these simple physics problems cause lots of problems in math classes.

Newton's Second Law for Springs gives -kx = ma.

In terms of differential equations m d²x/dt² + k x = 0.

This is a linear differential equation with constant coefficients. With the guess x = Ae^λt the characteristic equation becomes mλ² - k = 0. The solution for the characteristic is λ = ±i√k/m. For complex solutions to the characteristic equation λ = a ± i b (complex solutions always come in conjugate pairs),

x = A e^atsin(bt) + B e^at cos(bt).

We then have the solution

x = A sin( √k/m t) + B cos( √k/m t)

Using the angle addition formula (A = C cos(φ), B = C sin(φ)), and the expectation the mass will act like a spring the solutions can become,

x = C sin( √k/m t + φ).

The initial conditions from the problem are, x(0) = 4/12 ft, v(0) = 3 ft/s, x_max = 1/2 ft.

(1) x(0) = C sin(φ) = 4/12 ft

v(t) = C √k/m cos( √k/m t + φ)

(2) v(0) = C √k/m cos(φ) = 3 ft/s

(3) x_max = C = 1/2 ft

(3) → (1) sin(φ) = 2/3

sin²(φ) + cos²(φ) = 1

cos(φ) = √5 /3 (technically there are two solutions after the square root that give different φ values. You just need to pick one of the solutions.)

(2) √k/m = 18/ √5

x = 1/2 ft sin( 18/√5 t + arcsin(2/3))

Check

v(t) = 9/√5 ft cos( 18/√5 t + arcsin(2/3))

v(0) = 9/√5 ft cos( arcsin(2/3)) = 3