What is the phase difference for two particles executing simple harmonic motion?

[LandOfStandar]

[SOLVED] phase difference problem - need help

Homework Statement

Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. The pass each other moving in opposite directions each time their displacement is half their amplitude. What is the phase difference.

Homework Equations

X = A cos(ωt + θ)
T = 2π/ω
T = 1/f

The Attempt at a Solution

A is the same for both
f is the same, so t is the same
T is the same, so ω is the same
but since no numbers are given I am lost

 

[Mindscrape]

Think about superposition. One wave plus another wave. Do you know about euler's formula? Taking this into polar form will make it a lot easier.

 

[LandOfStandar]

no idea

 

[rootX]

Use the trig identity
cos A + cos B = 2 cos((A+B)/2)cos ((A-B)/2)
Your book should provide you a formula for superstition of waves.
For simplicity assume that one particles has a phase constant of 0.

 

[LandOfStandar]

I just got to this in wave, but the question was asked before this was taught. Is there any other way? If not, I will see if something like this was mentioned before that I missed.

 

[Gear300]

The two waves are parallel and meet at half the amplitude. That means that one of the waves could be above the other. While one of the waves is going under its equilibrium axis, the other is going above its equilibrium axis. The waves only meet at half the amplitude, so when you have a maximum of one wave, at the same time, the other wave is at a minimum. In that case, one wave is the negative of the other. There is no need to find the distance between the two waves, all that's needed is to find a phase constant that makes one of the functions the negative of the other.

 

[LandOfStandar]

I figure that, but can not figure how to find the phase constant with no numbers.

 

[Gear300]

No need for numbers. One wave function is simply negative to other (while ignoring the distance between the two waves); just find a phase constant for cos(x + C) that is analogous to -cos(x)

 

[Antenna Guy]

There is a number provided, and that number is [itex]\frac{1}{2}[/itex].

Where the particles pass one-another, their displacements are equal and the first derivatives of their displacement functions differ only in sign.

Regards,

Bill

 

[LandOfStandar]

X(1) = A cos(ωt), (1/2)A = A cos(ωt), (1/2)=cos(ωt), [cos arc(1/2)]/ω = t
X(2) = -A cos(ωt + θ), (1/2)A = -A cos(ωt + θ), [-cos arc(1/2)] = [ω[cos arc(1/2)]/ω+ θ],
[-cos arc(1/2)] = [cos arc(1/2)+ θ], [-cos arc(1/2)] - cos arc(1/2) = θ, -2cos arc(1/2) = 2.09 rads

Would that be right, yes or no? If no Where did I go wrong.

 

[Antenna Guy]

X(1) = A cos(ωt), (1/2)A = A cos(ωt), (1/2)=cos(ωt), [cos arc(1/2)]/ω = t
X(2) = -A cos(ωt + θ), (1/2)A = -A cos(ωt + θ), [-cos arc(1/2)] = [ω[cos arc(1/2)]/ω+ θ],
[-cos arc(1/2)] = [cos arc(1/2)+ θ], [-cos arc(1/2)] - cos arc(1/2) = θ, -2cos arc(1/2) = 2.09 rads

Would that be right, yes or no? If no Where did I go wrong.

not right. clues:

1) X(1)=X(2) since they are at the same height.
2) (ωt), and (ωt + θ) are the phases - you need θ
3) don't change the sign of the amplitude going from X(1) to X(2)

Regards,

Bill

 

[LandOfStandar]

Acos(ωt)=Acos(ωt + θ)
cos(ωt)=cos(ωt + θ)

where does the half come in, I am lost with this

 

[Antenna Guy]

Acos(ωt)=Acos(ωt + θ)
cos(ωt)=cos(ωt + θ)

where does the half come in, I am lost with this

What you did for X(1) is where the 1/2 comes in. Solve for ωt rather than t.

cos(ωt)=cos(ωt + θ)=1/2

Hint: for θ=0, X(1) and X(2) are going in the same direction...

Regards,

Bill

 

[LandOfStandar]

(1/2)A = A cos(ωt); (1/2) = cos(ωt); cos-1 (1/2) = (ωt)
(1/2)A = A cos(ωt+ θ); (1/2) = cos[(cos-1 (1/2))+ θ]; cos-1(1/2) - (cos-1 (1/2)) = θ;
0 = θ

ok I see your last hint, but if I don't change the sign for amplitude, what dictates it is going in the opposite direction?

 

[Antenna Guy]

(1/2)A = A cos(ωt); (1/2) = cos(ωt); cos-1 (1/2) = (ωt)
(1/2)A = A cos(ωt+ θ); (1/2) = cos[(cos-1 (1/2))+ θ]; cos-1(1/2) - (cos-1 (1/2)) = θ;
0 = θ

ok I see your last hint, but if I don't change the sign for amplitude, what dictates it is going in the opposite direction?

cos(ωt)=1/2 *twice* in a period (T).

If you are familiar with derivatives:

x=cos(ωt) <- position

dx/dt=-sin(ωt) <- velocity

In your problem, the *velocities* have different signs - not the amplitudes.

You can probably get the answer without doing derivtives, but that is where the answer comes from.

Regards,

Bill

 

[LandOfStandar]

I don't mean to be slow but I am lost:

Acos(ωt)=Acos(ωt + θ)
cos(ωt)=cos(ωt + θ)

where does the half come in, I am lost with this

X(1)=X(2) since they are at the same height.

cos(ωt)=cos(ωt + θ)

What you did for X(1) is where the 1/2 comes in. Solve for ωt rather than t.

(1/2)A = A cos(ωt); (1/2) = cos(ωt); cos-1 (1/2) = (ωt)

Put cos-1(1/2) in for (ωt) and solved
cos(cos-1 (1/2))=cos(cos-1 (1/2) + θ)
cos-1[cos(cos-1 (1/2))] = cos-1 (1/2) + θ
cos-1[cos(cos-1 (1/2))] - cos-1 (1/2) = θ = 0

What are you saying in the last post, held to bed but I will check for a post in the morning, thanks for working with me, I don't know why I am not getting this.

 

[Antenna Guy]

What are you saying in the last post, held to bed but I will check for a post in the morning, thanks for working with me, I don't know why I am not getting this.

Suggestion: Draw a cosine function from 0-T (i.e. X(1) versus ωt). Draw a horizontal line from 0-T at x=1/2. The first intersection is the phase of X(1) (compare with what you calculated), and the second intersection is the phase of X(2).

If you get that far without too much trouble, add a curve for -sin(ωt) and see how that matches up at the two points where X(1)=1/2.

I assume you don't quite know what to do with derivatives yet, but you can still get there. Compare how X(1) is changing at the two points - one should be going "up" while the other is going "down".

BTW - If you take a copy of your X(1) curve and slide it back to where the second point of the copy overlays the first point of the original, you will have overlayed both X(1) and X(2) as a function of t.

Keep asking if you still have questions.

Regards,

Bill

 

[LandOfStandar]

x(1)=cos(ωt)
x(2)=cos(ωt + θ)
x=(1/2)
x=x(1)=x(2)

cos(ωt)=(1/2)
(ωt)=cos-1(1/2)=Pi/3 or 2(pi)-Pi/3 = 5(pi)/3
(ωt)=Pi/3
(ωt + θ)=5(pi)/3, [Pi/3 + θ]=5(pi)/3, θ=4(pi)/3

cos(ωt)
cos[ωt + (4(pi)/3)]

They will not intersect every time at half amplitudes, but every other time at (1/2) and (-1/2) amplitudes.

Is this right even though it does not cross every time?

 

[Antenna Guy]

x(1)=cos(ωt)
x(2)=cos(ωt + θ)
x=(1/2)
x=x(1)=x(2)

cos(ωt)=(1/2)
(ωt)=cos-1(1/2)=Pi/3 or 2(pi)-Pi/3 = 5(pi)/3
(ωt)=Pi/3
(ωt + θ)=5(pi)/3, [Pi/3 + θ]=5(pi)/3, θ=4(pi)/3

cos(ωt)
cos[ωt + (4(pi)/3)]

They will not intersect every time at half amplitudes, but every other time at (1/2) and (-1/2) amplitudes.

Is this right even though it does not cross every time?

It looks good to me.

A more intuitive way to look at this problem (for me at least) would be to place 2 balls on the edge of a disk separated counter-clockwise by 240deg of arc (4(pi)/3 radians). Place one ball at the top of the disk and call it X(1). Now spin the disk clockwise at ω Hz.

If you look at the spinning disk edge-on, it should look just like the problem you solved.

A paper plate marked as described above, with a pencil poked through the center (so you can spin it in slo-mo) should be enough to see where the marks line up (and where they don't).

Regards,

Bill

P.S. Don't forget you can add + /- n*2(pi) to θ (where n is any integer).

 

[LandOfStandar]

thanks so much