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m
= (0, 0 , −g).
In other words, acceleration ¨q is a constant. Since there’s no increase in difficulty, we’ll just solve all constant acceleration problems in any dimension, i.e. differential equations of the form
q¨ = a
where a ∈ Rn^ is a constant. Then we’ll set n = 3 and a = (0, 0 , −g) to solve this special case. It’s straightforward to integrate q¨ = a
once and get q˙ = v + ta
where our constant of integration v ∈ Rn^ is seen, upon setting t = 0, to be the initial velocity
q˙(0) = v
so we have q˙ = ˙q(0) + ta
thus far, and we integrate again to get
q = x + t q˙(0) +
t^2 a
where, just as before, our constant of integration x ∈ Rn^ is seen, upon setting t = 0, to be the initial positon q(0) = x
So our full solution is thus
q = q(0) + t q˙(0) +
t^2 a
Now back to our original problem, where n = 3 and a = (0, 0 , −g). Taking the initial position and velocity to be q(0) = (x 0 , y 0 , z 0 )
and q˙(0) = (vx, vy , vz )
we get that our position at time t is
q(t) = (xo + vxt, yo + vy t, zo + vz t −
gt^2 )
in terms of its components. In particular, if we take the z-axis to point in the vertical direction, we have that the particle’s height at time t is
qz (t) = zo + vz t −
gt^2 ,
the familiar formula from any freshman physics text.
Homework 2
Solve Newton’s second law for q(t) ∈ R when the force is given by Hooke’s law
F (t) = −kq(t)
in terms of m, k, q(0), and ˙q(0), and then find the period P of the oscillation and the frequency ω,
where ω =
2 π P
Solution
Plugging this force into Newton’s second law, we get
q¨ = −
k m
q
or
q¨ + k m
q = 0
a slightly more difficult differential equation than the one in the first problem. To solve it, we turn to the arcane techniques of the theory of ordinary differential equations: we find the characteristic polynomial of this equation,
λ^2 + k m
which is the polynomial that looks like the differential equation, if we replace each nth derivative of q with the nth power of λ. Now we find the roots of this polynomial,
λ = ±i
k m
= ±iω
where we’ve defined ω =
k m. Since the characteristic polynomial has no repeated roots, the most general complex-valued solution to our differential equation is of the form
q˜(t) = Aeiωt^ + Be−iωt
where A and B are complex constants to be determined by initial conditions. But we don’t want a complex-valued solution; we want a real solution. This is easy to obtain, because the real part of ˜q,
q(t) = Re(˜q(t) = A′^ cos(ωt) + B′^ sin(ωt)
