Simple harmonic motion of particles

[erisedk]

Homework Statement

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 answer is π/2 but the answer given is π/3.

Homework Equations

x = Asin(ωt + φ)

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.

 

[ehild]

Homework Statement

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 answer is π/2 but the answer given is π/3.

Homework Equations

x = Asin(ωt + φ)

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.

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?

 

[BvU]

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

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 ...

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

 

[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.

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?

I think I'm very confused here. I don't know how to.

 

[ehild]

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

$\dfrac{1}{2} = cos\dfrac{2ωt + φ_1 + φ_2}{2}sin\dfrac{φ_2 - φ_1}{2}$
I don't know what to do beyond this.

I think I'm very confused here. I don't know how to.

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?

 

[erisedk]

Oh, I got that, I edited my post.

 

[ehild]

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

 

[erisedk]

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

 

[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.

 

[erisedk]

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