# Simple harmonic motion of particles

1. Mar 28, 2016

### erisedk

1. The problem statement, all variables and given/known data
Two particles are executing simple harmonic motion of the same amplitude A and frequency ω along the x-axis. Their mean position is separated by distance X0 (X0 > A). If the maximum separation between them is (X0 + A), the phase difference between their motion is

My

2. Relevant equations
x = Asin(ωt + φ)

3. The attempt at a solution
To me it looks like this: When the first particle is at it's mean position, the second particle is at it's extreme position, so that the distance between both of them is X0 + A. So, the phase difference should be π/2

In terms of coordinates, let mean pos of P1 be x=0, mean pos of P2 be X0. Let P1 go from x= -A to x = +A and P2 go from x = X0 - A to x = X0 + A. The situation here represents P1 at x=0 and P2 at x = X0 + A.

2. Mar 28, 2016

### ehild

You can check the validity of your logic, by finding the maximum separation if the phase difference is pi/2. Is it really xo+A?

3. Mar 28, 2016

### BvU

In that case A isn't the maximum difference ! Work out $A \sin \omega t (t + \phi) - A \sin(\omega t)$ to get an expression in $\phi$ and see where you can go from there ....

 ah! Elisabeth was faster. I leave you in her hands...

4. Mar 29, 2016

### erisedk

x1 = Asin(ωt + Φ1)
x2 = X0 + Asin(ωt + Φ2)
x2 - x1 = X0 + Asin(ωt + Φ2) - Asin(ωt + Φ1)
= X0 + 2A$cos\dfrac{2ωt + φ_1 + φ_2}{2}sin\dfrac{φ_2 - φ_1}{2}$
X0 + A = X0 + 2A$cos\dfrac{2ωt + φ_1 + φ_2}{2}sin\dfrac{φ_2 - φ_1}{2}$

$\dfrac{1}{2} = cos\dfrac{2ωt + φ_1 + φ_2}{2}sin\dfrac{φ_2 - φ_1}{2}$

This can be visualised as SHM with amplitude $sin\dfrac{φ_2 - φ_1}{2}$

$\dfrac{1}{2} = sin\dfrac{φ_2 - φ_1}{2}$

$φ_2 - φ_1 = \dfrac{π}{3}$

While I have gotten the answer, I'm not sure I understand why it's true at all.
I think I'm very confused here. I don't know how to.

Last edited: Mar 29, 2016
5. Mar 29, 2016

### ehild

You can choose one of the phase constant zero.
x1 = Asin(ωt )
x2 = X0 + Asin(ωt + φ)
x2 - x1 = X0 + Asin(ωt + φ) - Asin(ωt )
$cos\dfrac{2ωt + φ}{2}sin\dfrac{φ}{2}$
At what time is the last expression maximum and what is the maximum value?

6. Mar 29, 2016

### erisedk

Oh, I got that, I edited my post.

7. Mar 29, 2016

### ehild

It is confusing to the reader that you edited the first post. This makes to look all helpers were stupid.

8. Mar 29, 2016

### erisedk

I didn't edit my first post. I edited my answer post before you replied.

9. Mar 29, 2016

### ehild

Yes, I see. So I was stupid only, answering a solved problem.
Anyway. If you do correct Maths and you got a result it is correct.

10. Mar 29, 2016

### erisedk

I'm really sorry if I shouldn't have.