Please can I get some help finding the function of motion

[Alexan]

Homework Statement

find the function of motion

Relevant Equations

none

Homework Statement:: find the function of motion
Homework Equations:: none

i could find the amplitude and the phase angle but i can't find the phase difference and the function of motion.

 

[Delta2]

First of all the statement asks you to find amplitude ($A$), phase($\phi$) and Period ($T$). Not the phase difference (?)
The function of motion is to "combine" those three constants into the equation $$x(t)=A\cos(\omega t+\phi)$$ where $\omega=\frac{2\pi}{T}$

 

[Alexan]

you need the phase difference to get the angular frequency of the oscillation which will help you get the function of motion.
the answer you gave to me is wrong

 

[PeroK]

you need the phase difference to get the angular frequency of the oscillation which will help you get the function of motion.
the answer you gave to me is wrong

What @Delta2 said is correct.

 

[Delta2]

yes well, I think I understand now what you meant by phase difference and how useful it is to find angular frequency.

What is the total phase call it $\phi_0$ at time $t_0=0$?
What is the total phase call it $\phi_1$ at time $t_1=1$?

You can answer these two questions by just looking at the diagram.

After that you can compute the angular frequency from the equation $\phi_1-\phi_0=\omega (t_1-t_0)$ is that what you meant?

 

[Alexan]

yes well, I think I understand now what you meant by phase difference and how useful it is to find angular frequency.

What is the total phase call it $\phi_0$ at time $t_0=0$?
What is the total phase call it $\phi_1$ at time $t_1=1$?

You can answer these two questions by just looking at the diagram.

After that you can compute the angular frequency from the equation $\phi_1-\phi_0=\omega (t_1-t_0)$ is that what you meant?

Exactly now my issue is that i don't know how to find it from the diagram.

 

[Alexan]

What @Delta2 said is correct.

The issue is that i need to get the phase difference from the diagram which is a little bit difficult for me

 

[PeroK]

The issue is that i need to get the phase difference from the diagram which is a little bit difficult for me

You don't need to calculate a phase difference. You can solve the question more easily without it.

1) You got $A$ from the maximum height of the wave, I imagine.

2) Using $t = 0$ gives you $\phi$.

3) Using $t = 1$ gives you $\omega$, hence $T$.

PS perhaps using $x(t) = A \sin(\omega t + \phi)$ might be simpler in this case.

 

[Alexan]

You don't need to calculate a phase difference. You can solve the question more easily without it.

1) You got $A$ from the maximum height of the wave, I imagine.

2) Using $t = 0$ gives you $\phi$.

3) Using $x = 1$ gives you $\omega$, hence $T$.

PS perhaps using $x(t) = A \sin(\omega t + \phi)$ might be simpler in this case.

Thanks for your reply i can't get your 2nd and 3rd point can you write it down and send me the picture or just give a detail explanation.

 

[Delta2]

Exactly now my issue is that i don't know how to find it from the diagram.

You said that you found the phase angle, isn't that equal to the total phase $\phi_0$ at time $t_0=0$?
As for the phase $\phi_1$, what is the phase when the graph of x(t) crosses the x-axis, that is when x(t)=0?

 

[Alexan]

Φ

You said that you found the phase angle, isn't that equal to the total phase $\phi_0$ at time $t_0=0$?
As for the phase $\phi_1$, what is the phase when the graph of x(t) crosses the x-axis, that is when x(t)=0?

The phase angle i got was from cos(Φ)=initial displacement/amplitude=3/6 from the oscillating curve

 

[PeroK]

Thanks for your reply i can't get your 2nd and 3rd point can you write it down and send me the picture or just give a detail explanation.

I'll hand over to @Delta2 as there's no point both of us trying to help you.

 

[Alexan]

You said that you found the phase angle, isn't that equal to the total phase $\phi_0$ at time $t_0=0$?
As for the phase $\phi_1$, what is the phase when the graph of x(t) crosses the x-axis, that is when x(t)=0?

What you say is correct but yet i don't know how to get it from the above diagram.

 

[Alexan]

I'll hand over to @Delta2 as there's no point both of us trying to help you.

There is, i would like to get more explanation from both of you.

 

[Delta2]

Well first of all as @PeroK said, it would be better to use $x(t)=A\sin(\omega t+\phi)$ in this case cause we can easily see from the diagram that is the graph of $\sin x$ shifted by a small phase $0<\phi<\frac{\pi}{2}$.

After that , don't we know that when the graph crosses the x-axis the phase can be $0$ or $\pi$ or $2\pi$ or in general $k\pi$ where k a positive integer. What is the value of k for the case where the graph crosses the x-axis for the first time?

 

[Alexan]

Well first of all as @PeroK said, it would be better to use $x(t)=A\sin(\omega t+\phi)$ in this case cause we can easily see from the diagram that is the graph of $\sin x$ shifted by a small phase $0<\phi<\frac{\pi}{2}$.

After that , don't we know that when the graph crosses the x-axis the phase can be $0$ or $\pi$ or $2\pi$ or in general $k\pi$ where k a positive integer. What is the value of k for the case where the graph crosses the x-axis for the first time?

This is not accurate. what about this kind of oscillation.

 

[Alexan]

This is not accurate. what about this kind of oscillation.View attachment 253858

i think you have to solve it normally and ignore the fact that it is a sine graph

 

[Delta2]

Ok well the problem to find the phase angle is that when we going to solve for the equation $\sin\phi=x(0)$ or $\cos\phi=x(0)$ we got to know where about is $\phi$ more specifically whether it is $0<\phi<\pi$ or $\pi<\phi<2\pi$ otherwise we can't be sure what is the value of $\phi$.

 

[Alexan]

Ok well the problem to find the phase angle is that when we going to solve for the equation $\sin\phi=x(0)$ or $\cos\phi=x(0)$ we got to know where about is $\phi$ more specifically whether it is $0<\phi<\pi$ or $\pi<\phi<2\pi$ otherwise we can't be sure what is the value of $\phi$.

OK let's say the standard conditions are
π<ϕ<π

 

[Delta2]

Then you can uniquely determine the phase angle in the case of the equation $\cos\phi=x_0$ otherwise you cannot.

 

[Alexan]

ok

Then you can uniquely determine the phase angle in the case of the equation $\cos\phi=x_0$ otherwise you cannot.

 

[Delta2]

Back at the original exercise , what do you think is the phase $\phi_1$ when the graph crosses the x-axis? Is it 0, $\pi$ or $2\pi$??.

 

[Alexan]

Back at the original exercise , what do you think is the phase $\phi_1$ when the graph crosses the x-axis? Is it 0, $\pi$ or $2\pi$??.

to be sincere with you i don't know because the x-axis is in terms of time.

 

[PeroK]

to be sincere with you i don't know because the x-axis is in terms of time.

Back to the question:

The horizontal axis is the t-axis; the vertical axis in the x-axis. If you are happier with a cosine function, you can stick with that.

What do you about the system at time $t = 0$?

 

[Alexan]

Back to the question:

The horizontal axis is the t-axis; the vertical axis in the x-axis. If you are happier with a cosine function, you can stick with that.

What do you about the system at time $t = 0$?

thanks for your assistance but i think we are getting out of phase.
If you could send me the mathematical expression for what your saying it will help me more understand

 

[PeroK]

thanks for your assistance but i think we are getting out of phase.
If you could send me the mathematical expression for what your saying it will help me more understand

Here's the graph.

Here's the equation of that graph:

$$x(t)=A\cos(\omega t+\phi)$$ where $\omega=\frac{2\pi}{T}$

 

[Delta2]

Ok I ll try to explain this as more mathematically as I can.

Lets start with the equation $x(t)=A\sin(\omega t+\phi)$ for which we are given the graph.
Then do the change of variable $\Phi=\omega t+\phi$ so the equation becomes

$x(\Phi)=A\sin\Phi$.
If i ask you at what times the first equation crosses the x-axis you ll tell me at t=1 and t=2.
If i ask you at what angles $\Phi$ the second equation crosses the x-axis what will you answer to me?

 

[Alexan]

Here's the graph.

View attachment 253860
Here's the equation of that graph:

according to you what is the period T

 

[PeroK]

according to you what is the period T

$T = \frac{2\pi}{\omega}$

 

[Alexan]

Ok I ll try to explain this as more mathematically as I can.

Lets start with the equation $x(t)=A\sin(\omega t+\phi)$ for which we are given the graph.
Then do the change of variable $\Phi=\omega t+\phi$ so the equation becomes

$x(\Phi)=A\sin\Phi$.
If i ask you at what times the first equation crosses the x-axis you ll tell me at t=1 and t=2.
If i ask you at what angles $\Phi$ the second equation crosses the x-axis what will you answer to me?

Thanks for your reply

 

[Alexan]

$T = \frac{2\pi}{\omega}$

?

 

[PeroK]

?

I assume that your book has a description and analysis of simple harmonic motion. I think you need to re-read that section.

If not, try:

https://www.khanacademy.org/science...introduction-to-simple-harmonic-motion-review