There is no phase angle here as the coefficient of sin (6t) = 0
x= Acos(6t)+Bsin(6t) + 5/27 cos(3t)
x(0) = A + 5/27 = 0 A = -5/27
x ' (0) = 6B = 0 B = 0
x(t) = -5/27 cos(6t) +5/27cos(3t)
Hi, For the following problem I am having hard time trying to find the phase angle (alpha).
How to find the phase angle,
isnt that alpha = arctan(B/A) if A,B>0 ?
Problem:
Solve the initial value problem
(where primes indicate derivatives with respect to t). Express your solution as the sum of two oscillations,
x(t) = (49/27)*cos(6t - ) + (5/27)cos(3t)
where =
(Recall that must be between 0 and 2.)
But the phase angle is not right.
Please help,
Many Thanks.