Phase Difference in Parallel SHM with Equal Amplitude and Frequency

[EddiePhys]

Homework Statement

Two particles move parallel to the x-axis about the origin with the same amplitude and frequency. At a certain instant, they are found at a distance A/3 from the origin, on opposite sides of the origin, with their velocities in the same direction. Find the phase difference.

Homework Equations

Phasor diagram, with the projection on the y-axis executing shm

The Attempt at a Solution

What have I done wrong?

 

[hilbert2]

Let's say the trajectories are $x_1 (t)$ and $x_2 (t)$. The general solution is

$x_1 (t) = A\sin (\omega t + \delta_1 )$
$x_2 (t) = A\sin (\omega t + \delta_2 )$

where $A$ is the common amplitude and $\omega$ is the common frequency. Now, due to the arbitrariness in choosing the moment of time where $t=0$, you can decide that $\delta_2 = 0$ (try to convince yourself of the fact that this is allowed). Also, you can divide the $x_1$ and $x_2$ with the ampitude $A$, obtaining scaled coordinates $\tilde{x}_{1}$ and $\tilde{x}_{2}$ where the amplitude of the motion is $1$. Now, how to invert a sine function?

Edit: Also note that the state of a classical mechanical system is described by giving both positions and velocities, so you also have equations for $\frac{dx_1}{dt}$, etc..

 

[Merlin3189]

You started ok with your circle and phasors, but why introduce angle α ? (There is also another position the phasors could be.)
What is the relationship between θ and α ?

I agree with your calculation for θ and your calculation for α.

I agree with your answer for the phase difference, though I chose the smaller phase angle.

Since there is no indication of which phase is required, you can simply look at your diagram and choose the easiest angle to calculate.

 

[EddiePhys]

You started ok with your circle and phasors, but why introduce angle α ? (There is also another position the phasors could be.)
What is the relationship between θ and α ?I agree with your calculation for θ and your calculation for α.

I agree with your answer for the phase difference, though I chose the smaller phase angle.

Since there is no indication of which phase is required, you can simply look at your diagram and choose the easiest angle to calculate.

Since they both have the same velocity direction (given in the question) and the both phasor particles move anticlockwise, the positions I chose for the two particles are the only two positions such that their projections are moving in the same direction with equal and opposite distance from the origin

 

[EddiePhys]

Let's say the trajectories are $x_1 (t)$ and $x_2 (t)$. The general solution is

$x_1 (t) = A\sin (\omega t + \delta_1 )$
$x_2 (t) = A\sin (\omega t + \delta_2 )$

where $A$ is the common amplitude and $\omega$ is the common frequency. Now, due to the arbitrariness in choosing the moment of time where $t=0$, you can decide that $\delta_2 = 0$ (try to convince yourself of the fact that this is allowed). Also, you can divide the $x_1$ and $x_2$ with the ampitude $A$, obtaining scaled coordinates $\tilde{x}_{1}$ and $\tilde{x}_{2}$ where the amplitude of the motion is $1$. Now, how to invert a sine function?

Edit: Also note that the state of a classical mechanical system is described by giving both positions and velocities, so you also have equations for $\frac{dx_1}{dt}$, etc..

I get this, but why is my solution wrong?

 

[hilbert2]

I get this, but why is my solution wrong?

I can't see you taking the velocity functions $\frac{dx_1 (t)}{dt} = A\omega \cos (\omega t + \delta)$ and the same for $x_2$ in account anywhere in the calculation.

 

[EddiePhys]

I can't see you taking the velocity functions $\frac{dx_1 (t)}{dt} = A\omega \cos (\omega t + \delta)$ and the same for $x_2$ in account anywhere in the calculation.

I did, that's how I got my phasor diagrams. Since both the particles move anticlockwise, the position I've chosen is the only one where the two projections both have the same direction of velocity as well as equal and opposite displacements from the origin

 

[hilbert2]

You should write the problem explicitly as a system of equations for the positions and velocities, and then eliminate the parameters $A$, $t$ and $\omega$ so that only the phase $\delta$ remains.

 

[EddiePhys]

You should write the problem explicitly as a system of equations for the positions and velocities, and then eliminate the parameters $A$, $t$ and $\omega$ so that only the phase $\delta$ remains.

I understand that method. I just want to know why what I have done is wrong

 

[TSny]

Since they both have the same velocity direction (given in the question) and the both phasor particles move anticlockwise, the positions I chose for the two particles are the only two positions such that their projections are moving in the same direction with equal and opposite distance from the origin

Your diagram does not correspond to the particles having the same sign of v_x, nor does it correspond to the particles having opposite signs for x.
[EDIT: Your diagram is OK. I mistakenly assumed that the horizontal axis in your phasor diagram corresponded to the x axis. Instead, the x-axis must be the vertical axis (which I should have seen from the fact that your markings of A/3 are on the vertical axis). Sorry for bungling this.]

 

[Merlin3189]

Since they both have the same velocity direction (given in the question) and the both phasor particles move anticlockwise, the positions I chose for the two particles are the only two positions such that their projections are moving in the same direction with equal and opposite distance from the origin

I just don't understand why people are making it so difficult. It seems to me you've solved it when you get sin θ = 1/3

 

[TSny]

View attachment 206854

I just don't understand why people are making it so difficult. It seems to me you've solved it when you get sin θ = 1/3

Yes. I now see what my problem was. I was taking the horizontal axis as being the x-axis in EddiePhys's diagram. I should have looked at it more carefully. It is OK if you take the vertical axis as the x-axis as you have shown in your diagram. I agree with you. Sorry for adding to the confusion.

 

[EddiePhys]

View attachment 206854

I just don't understand why people are making it so difficult. It seems to me you've solved it when you get sin θ = 1/3

Oh okay, I got it. Though not mentioned, they were probably expecting the 2nd diagram(green arrows) with the smaller phase angle difference. Thanks a lot! :D

 

[Merlin3189]

Yes, I agree I'd go for the smaller angle. I just mentioned it because in some situations it's important to be aware that there may be several solutions.

I'm also not sure whether you've seen the symmetry here and that α = π/2 - θ

So what is your final answer?