Understanding Simple Harmonic Motion: Explaining x=Acos(wt+phi)

[Jozefina Gramatikova]

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

 

[PeroK]

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

It depends where the system starts at $t=0$.

 

[Jozefina Gramatikova]

So, if it starts from x=0 at t=0 => phi=0? and at t=0 x=1 => phi would be different from zero?

 

[PeroK]

So, if it starts from x=0 at t=0 phi=0? and at t=0 x=1 phi would be different from zero?

Yes. $-A \le x(0) \le A$, where $x(0)$ is the displacement of the system at $t=0$, and determines $\phi$.

 

[vela]

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?

 

[Jozefina Gramatikova]

Plug t=0 and $\phi=0$ into $x(t) = A\cos(\omega t + \phi)$. Is $x(0)$ equal to 0?

x(t)=Acos(0)
x(t)=A

 

[vela]

x(t)=Acos(0)
x(t)=A

Presumably $A \ne 0$ otherwise there'd be no oscillating going on, so clearly $x(0)=0$ and $\phi=0$ contradict each other.

 

[haruspex]

So, if it starts from x=0 at t=0 => phi=0?

No.
If it is at x₀ at t=0 and at x=Acos(ωt+φ) at time t, what equation can you write for x₀?

 

[Jozefina Gramatikova]

I don't know I am really confused right now.

 

[PeroK]

I don't know I am really confused right now.

$\cos(0) = 1$

 

[Jozefina Gramatikova]

$\cos(0) = 1$

I know that :D Thats why I said:
"x(t)=Acos(0)
x(t)=A"

 

[haruspex]

I don't know I am really confused right now.

Just substitute those t and x values in the general equation in post #8.

 

[haruspex]

x(t)=Acos(0)

When plugging in a value for t you should replace all occurrences of t.

 

[haruspex]

I know that :D Thats why I said:
"x(t)=Acos(0)
x(t)=A"

You did not seem to know it in post #3.

 

[Jozefina Gramatikova]

Just substitute those t and x values in the general equation in post #8.

x=Acos(ωt+φ)
x0=Acos(ω x 0+φ)
x0=Acos(ω x 0+φ)
x0=Acos(φ)?

 

[Jozefina Gramatikova]

Could you give me the right solution, please? I get more and more confused with every single post.

 

[Ray Vickson]

Could you give me the right solution, please? I get more and more confused with every single post.

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 $t$.

If $\phi = 0$ we have $x(0) = A$, so that means that the motion starts at time zero from the upper end of the allowed interval. Then for a while $x(t)$ will decrease until it hits the lower end $-A$ of the allowed interval, at time $\pi/\omega$. Then it will start to increase until it hits $x=A$ at time $2 \pi/\omega,$ etc., etc.

If you want $x(0)$ to be a point other than $A$ you need to have $\phi \neq 0.$ To help you to enhance your understanding, try the following question: "what would be the value of $\phi$ if you require $x(0) = 0?$ (Of course, because the $\cos$ function is periodic there will be many values of $\phi$ that will work, but among these there is one that is "simplest", and that is the one you should attempt to find.)

 

[haruspex]

x0=Acos(φ)?

Yes.