- #1

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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)

- Thread starter Jozefina Gramatikova
- Start date

- #1

- 64

- 9

x=Acos(wt+phi)

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)

- #2

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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

- #3

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

- #4

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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?

- #5

vela

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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?

- #6

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

x(t)=A

- #7

vela

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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

- #8

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No.So, if it starts from x=0 at t=0 => phi=0?

If it is at x

- #9

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

- #10

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##\cos(0) = 1##I don't know I am really confused right now.

- #11

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- #12

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Just substitute those t and x values in the general equation in post #8.I don't know I am really confused right now.

- #13

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When plugging in a value for t you should replace all occurrences of t.x(t)=Acos(0)

- #14

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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"

- #15

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x=Acos(ωt+φ)Just substitute those t and x values in the general equation in post #8.

x0=Acos(ω x 0+φ)

x0=Acos(ω x 0+φ)

x0=Acos(φ)?

- #16

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

- #17

Ray Vickson

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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.

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.)

Last edited:

- #18

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Yes.x0=Acos(φ)?

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