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Homework Statement
x=Acos(wt+phi)
Homework Equations
can somebody explain to me please when phi=0. I saw many different questions with many solutions and I can't understand when we have just x=Acos(wt) and when x=Acos(wt+phi)
It depends where the system starts at ##t=0##.Homework Statement
x=Acos(wt+phi)
Homework Equations
can somebody explain to me please when phi=0. I saw many different questions with many solutions and I can't understand when we have just x=Acos(wt) and when x=Acos(wt+phi)
The Attempt at a Solution
Yes. ##-A \le x(0) \le A##, where ##x(0)## is the displacement of the system at ##t=0##, and determines ##\phi##.So, if it starts from x=0 at t=0 phi=0? and at t=0 x=1 phi would be different from zero?
Plug t=0 and ##\phi=0## into ##x(t) = A\cos(\omega t + \phi)##. Is ##x(0)## equal to 0?So, if it starts from x=0 at t=0 => phi=0? and at t=0 x=1 => phi would be different from zero?
x(t)=Acos(0)Plug t=0 and ##\phi=0## into ##x(t) = A\cos(\omega t + \phi)##. Is ##x(0)## equal to 0?
Presumably ##A \ne 0## otherwise there'd be no oscillating going on, so clearly ##x(0)=0## and ##\phi=0## contradict each other.x(t)=Acos(0)
x(t)=A
No.So, if it starts from x=0 at t=0 => phi=0?
##\cos(0) = 1##I don't know I am really confused right now.
Just substitute those t and x values in the general equation in post #8.I don't know I am really confused right now.
When plugging in a value for t you should replace all occurrences of t.x(t)=Acos(0)
You did not seem to know it in post #3.I know that :D Thats why I said:
"x(t)=Acos(0)
x(t)=A"
x=Acos(ωt+φ)Just substitute those t and x values in the general equation in post #8.
Let's assume that ##A>0##, so when ##x(t) = A \cos( \omega t + \phi)##, the value of ##x(t)## oscillates back and forth between ##-A## and ##+A##. That is, ##-A \leq x(t) \leq A## for all ##t##. Furthermore, we have ##x = -A## for some ##t## and ##x = +A## for some other ##t##.Could you give me the right solution, please? I get more and more confused with every single post.
Yes.x0=Acos(φ)?