286 Ordinary Differential Equations
The roots of the auxiliary equation, mr^2 + k = 0, are r 1 = ../klffii = wi
and r 2 = -../klffii = -wi. So the general solution of the differential
equation in (23) is either
where A 1 and A 2 are constants which are to be chosen to satisfy the
initial conditions or equivalently
y(t) = Asin(wt + ¢)
where A and¢ are constants which are to be chosen to satisfy the initial
conditions.
Differentiating the second form of the general solution of the differ-
ential equation, we see that
y'(t) = wAcos (wt+¢).
In order to satisfy the initial conditions given in (23), the constants A
and ¢ must be chosen to simultaneously satisfy
y(O) =A sin¢= -.6 rads and y'(O) = wAcos¢ = .3 rads/s.
Dividing the first equation by the second, we find ¢ must satisfy
sin¢ -.6
---= - = -2 s.
wcos¢ .3
Multiplying by w = ..Jk7ffi = )2.178/.2 = 3.3 rads/s, we see tan¢=
-6.6. So ¢ = -1.42 radians. Next, solving the equation A sin¢ = -.6
for A, yields
A= -.6 = -.6
sin¢ sin( -1.42)
-.6
__
98865
= .607 rads.
Thus, the equation of motion for this pendulum is
y(t) =A sin (wt+¢) = .607 sin (3.3t - 1.42).
b. The amplitude is A = .607 radians.
The period is P = 27r/w = 27r/3.3 = 1.904 seconds.
The frequency is F = 1/ P = .5252 cycles/second.
c. The phase angle is ¢ = -1.42 radians.