Write q(t) = (q1(t),q2(t), q3(t)) where qi:R→R. From Newton’s 2nd law, we have
¨q1(t) = ¨q2(t) = 0 and ¨q3(t) = −g. It follows that qi(t) = qi(0) + ˙qi(0)tfor i= 1,2 and q3(t) =
q3(0) + ˙q3(0)t−gt2/2. Therefore,
q(t) = (q1(0) + ˙q1(0)t, q2(0) + ˙q2(0)t, q3(0) + ˙q3(0)t−g
2t2).
Newton’s 2nd law and linear acceleration.
Consider a mass mattached to an ideal spring. Define the p osition qto be the distance of the mass
relative to the equlibrium position (i.e., when the system is at rest). By Hooke’s law, the force in
this situation is approximately given by
F(t) = −kq(t),
where k≥0 is the spring constant.
Solve Newton’s second law for q(t)∈Rin terms of m,k,q(0), ˙q(0) and find the period Pof
the oscillation and the frequency ω, where ω=2π
PHint: It oscillates!
Solution: Newton’s second law says
¨q(t) + k
mq(t) = 0
which has a general solution of
q(t) = Acos rk
mt!+Bsin rk
mt!.
With our initial conditions, we have
q(t) = q(0) cos rk
mt!+ ˙q(0)rm
ksin rk
mt!.
Note the period and frequency are P= 2πpm/k and ω=pk/m, respectively.
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