- #1
cosine
- 4
- 0
Hi everyone,
I ran into the following little problem:
Given v=40 cos(100*pi*t + pi/3), the question is:
what is the minimum number of millisec that the function must be shifted to the left if the expression for v is 40 cos(100*pi*t)
ok, I started with:
40 cos(100*pi*(t+t0) + pi/3) = 40 cos(100*pi*t)
hence 100*pi*(t+t0) + pi/3 = 100*pi*t
the 100*pi*t cancels out on each side and we can solve for t0 but t0 < 0
BUT the book starts with:
40 cos(100*pi*(t+t0) + pi/3) = 40 cos(100*pi*t + 2pi)
and they solved for t0... and t0 is now > 0
My question why they added the 2pi! Although the more I think about it the more it makes sense to me but I'd like to be sure... Is it because we look for the min time but still positive?
If that's the case, what bothers me is: how are we supposed to know ahead of time to add this '2pi'?
Thanks for your answer(s)
I ran into the following little problem:
Homework Statement
Given v=40 cos(100*pi*t + pi/3), the question is:
what is the minimum number of millisec that the function must be shifted to the left if the expression for v is 40 cos(100*pi*t)
Homework Equations
The Attempt at a Solution
ok, I started with:
40 cos(100*pi*(t+t0) + pi/3) = 40 cos(100*pi*t)
hence 100*pi*(t+t0) + pi/3 = 100*pi*t
the 100*pi*t cancels out on each side and we can solve for t0 but t0 < 0
BUT the book starts with:
40 cos(100*pi*(t+t0) + pi/3) = 40 cos(100*pi*t + 2pi)
and they solved for t0... and t0 is now > 0
My question why they added the 2pi! Although the more I think about it the more it makes sense to me but I'd like to be sure... Is it because we look for the min time but still positive?
If that's the case, what bothers me is: how are we supposed to know ahead of time to add this '2pi'?
Thanks for your answer(s)