At what angle will two pendulums collide?

  • #1

Homework Statement



Two masses, m1=750g and m2=500g, act as two pendulums with length L=2.0m. At t=0, they are released without an initial velocity, the former at -10° and the latter at 20°.
At what angle will they collide?

Homework Equations



θ(t)=Acos(wt)
sinθ=θ (small angle)
w=sqrt(g/l)

The Attempt at a Solution



I've found the frequency, which is 2.21 rad/s, and is identical for both masses. There are no phase constants as they are released from rest. So now I have:
θ1(t)=-0.175*cos(2.21t)
θ2(t)=0.349*cos(2.21t)

Now I can't just set them equal to each other, the cosines will cancel out and I'll be left with -.175=.349...

Sorry for not using LaTex, I really don't get it =/

Thank you,
 

Answers and Replies

  • #2
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Hmm, you're missing phase information. Every second order differential equation has 2 constants that determine the solution. You solved for amplitudes, but you need to apply initial conditions to get the right phases. Once you do that the cosines will not cancel out.
 
  • #3
PeterO
Homework Helper
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Homework Statement



Two masses, m1=750g and m2=500g, act as two pendulums with length L=2.0m. At t=0, they are released without an initial velocity, the former at -10° and the latter at 20°.
At what angle will they collide?

Homework Equations



θ(t)=Acos(wt)
sinθ=θ (small angle)
w=sqrt(g/l)

The Attempt at a Solution



I've found the frequency, which is 2.21 rad/s, and is identical for both masses. There are no phase constants as they are released from rest. So now I have:
θ1(t)=-0.175*cos(2.21t)
θ2(t)=0.349*cos(2.21t)

Now I can't just set them equal to each other, the cosines will cancel out and I'll be left with -.175=.349...

Sorry for not using LaTex, I really don't get it =/

Thank you,

I assume you are after the angle of the strings, relative to the vertical, when the masses meet.

The figure for frequency you calculated [in red above] does not include units of frequency - rad/s is a unit of angular velocity. Have you actually calculated frequency?

Have you ever seen a pendulum swing? Particularly have you ever seen a pendulum with large and small amplitude swing?

What are the conditions for a pendulum to display simple harmonic motion? What are the limits?
 
  • #4
Hmm, you're missing phase information. Every second order differential equation has 2 constants that determine the solution. You solved for amplitudes, but you need to apply initial conditions to get the right phases. Once you do that the cosines will not cancel out.

So would a phase consant of pi for one them solve the problem because it was released on the other side?

I assume you are after the angle of the strings, relative to the vertical, when the masses meet.

The figure for frequency you calculated [in red above] does not include units of frequency - rad/s is a unit of angular velocity. Have you actually calculated frequency?

Have you ever seen a pendulum swing? Particularly have you ever seen a pendulum with large and small amplitude swing?

What are the conditions for a pendulum to display simple harmonic motion? What are the limits?

My book says angular frequency IS in rad/s... And pendulums display SHM when theta is small; about 20° or less.
 
  • #5
PeterO
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So would a phase consant of pi for one them solve the problem because it was released on the other side?



My book says angular frequency IS in rad/s... And pendulums display SHM when theta is small; about 20° or less.

This question is simpler than you are making out.

Firstly - a pendulum has a fixed frequency/Period - that is why they can use them in a clock. Meaning is takes x seconds to complete a complete swing [out and back] - where x in the Period, or alternately completes f swings each second, where f is the frequency.

I didn't realise you were calculating angular frequency, since you only called it frequency. I am not sure why you wanted to calculate that angular frequency either.

The period of the pendulum is given by T = 2π√(l/g).

l is the same for both pendulums, and g is, of course, also equal.
What does that say about T?

Compared to the Period T, how long does it take for a pendulum to move from the extreme position [where they were both released from] to the mean position? [note, although released from different angles, those angles represent the extreme position for each]

What does that tell you?
 
  • #6
That they will hit exactly between the two starting positions?
 
  • #7
PeterO
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That they will hit exactly between the two starting positions?

Exactly!!! Provided you mean at the mean position - not 5 degrees to one side.

Quite simple wasn't it - especially when you consider the Physics and not the Maths.
 
  • #8
And let's assume I'd have to show the maths using all the formulae... would using a phase constant of pi give that answer? Just to make sure I understand phase constants corrrectly...
 
  • #9
PeterO
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Exactly!!! Provided you mean at the mean position - not 5 degrees to one side.

Quite simple wasn't it - especially when you consider the Physics and not the Maths.

I hope you saw my edit.
 
  • #10
Well exactly between both starting positions is +5° right?
 
  • #11
PeterO
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And let's assume I'd have to show the maths using all the formulae... would using a phase constant of pi give that answer? Just to make sure I understand phase constants corrrectly...

I suspect a phase constant of ∏ would be giving the correct answer - I have never analysed a pendulum problem that way.

If you imagine the two pendulums are set-up slightly off-set, so that they swing past each other rather than collide, then the graph of each would thus be a cos wave and a -cos wave. they would be equal at ∏/2, 3∏/2 etc despite the different amplitudes.
 
  • #12
PeterO
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Well exactly between both starting positions is +5° right?

That is not the correct answer.

The period of each pendulum is the same.
The time taken to get from extreme position to mean position is 1/4 of a period.

Each pendulum takes the same time to reach the mean position - so they collide at the 0o position.

remember these pendulums are not travelling at the same speed at any time, other than when they started with velocity zero.
 
  • #13
I understand... Thank you very much
 
  • #14
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Peter is right that you ultimately need to take a step back from everything and look at what you're doing. Still though, if you're going to just plug and chug you need to do it correctly. You should have had phase terms in there. If you had put the phase terms in there you would have realized that you had a transcendental equation, and that you must not be doing the problem right.

However, I don't know what either of you are talking about with a phase shift of -π.
θ1(t)=-0.175*cos(2.21t+Ø1)
apply the initial condition that at t=0 you need to be at -10*pi/180 radians, and you get a phase shift of .0721.

Hmm, how to explain phase shift... Phase shift is just a fancy way of saying how much the cosine, or sine, term has moved from its 0 phase position. We see a similar idea in say x^2 compared to (x-3)^2, the 3 shifted the x^2 graph to the right 3 units in space. If we have cos(θ-pi/2) all the values that used to start at cos(θ), the initial phase, have now shifted by -pi/2, so now we are at sin(θ). Compare the two graphs to see what I'm saying.
 
  • #15
Doesn't the phase shift formula have the initial velocity in it? Since it's zero I don't know how you managed to get 0.0721...
 
  • #16
PeterO
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Peter is right that you ultimately need to take a step back from everything and look at what you're doing. Still though, if you're going to just plug and chug you need to do it correctly. You should have had phase terms in there. If you had put the phase terms in there you would have realized that you had a transcendental equation, and that you must not be doing the problem right.

However, I don't know what either of you are talking about with a phase shift of -π.
θ1(t)=-0.175*cos(2.21t+Ø1)
apply the initial condition that at t=0 you need to be at -10*pi/180 radians, and you get a phase shift of .0721.

Hmm, how to explain phase shift... Phase shift is just a fancy way of saying how much the cosine, or sine, term has moved from its 0 phase position. We see a similar idea in say x^2 compared to (x-3)^2, the 3 shifted the x^2 graph to the right 3 units in space. If we have cos(θ-pi/2) all the values that used to start at cos(θ), the initial phase, have now shifted by -pi/2, so now we are at sin(θ). Compare the two graphs to see what I'm saying.

Now I am seeing a real problem with the angular frequency.

SHM - which a pendulum approximates.

SHM can be compared to a component of circular motion. That circular motion can have the "angular frequency" you are referring to.

Even wikipedia says "ω is the angular frequency or angular speed (measured in radians per second),"

If you wish analyse these pendulums using this "angular frequency" one pendulum is starting at 0, the other at ∏. Neither of them starts at -10*∏/180; that refers to an angle in the physical set-up - not an angle in the "circular motion equivalent" which the formulas analyse.

If you are starting at -10 and +20 degrees, then the speeds gradually increase until the masses collide when both strings are at 0 degrees. EDIT: this last sentence refers to the physical setup, not the mathematical formulae you are using.
 
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  • #17
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Is the location of these two masses hanging from a string somehow understood in this problem? Because I don't see it. If they are touching to start with it's easy because the period for the small arc is independent of mass and the angle and dependent on the length of the string and the planet but does it say they are touching before they are displaced?? Or is there a picture??
 
  • #18
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Now I am seeing a real problem with the angular frequency.

SHM - which a pendulum approximates.

SHM can be compared to a component of circular motion. That circular motion can have the "angular frequency" you are referring to.

Even wikipedia says "ω is the angular frequency or angular speed (measured in radians per second),"

If you wish analyse these pendulums using this "angular frequency" one pendulum is starting at 0, the other at ∏. Neither of them starts at -10*∏/180; that refers to an angle in the physical set-up - not an angle in the "circular motion equivalent" which the formulas analyse.

If you are starting at -10 and +20 degrees, then the speeds gradually increase until the masses collide when both strings are at 0 degrees. EDIT: this last sentence refers to the physical setup, not the mathematical formulae you are using.
Okay, well we agree on one thing, the masses will collide when they are at 0 degrees. But I have no idea what you're talking about with circular motion equivalent. Mass on spring has the same SHM as a pendulum (at least a pendulum under small angle approximation) and it doesn't move in a circle. The general solution to a second order differential equation x''+x=0 is x(t) or Ø(t) = A*cos(2*π*f+Phase). What I said about the phases for this problem is true, if you don't believe me solve for yourself and get the numerical answer to the transcendental.

Edit: Oh, I think maybe I see your point. As netgypsy says, there was no explicit reference of the angles to a vertical or anything, I was assuming the angles were with respect to the vertical.
 
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  • #19
PeterO
Homework Helper
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Is the location of these two masses hanging from a string somehow understood in this problem? Because I don't see it. If they are touching to start with it's easy because the period for the small arc is independent of mass and the angle and dependent on the length of the string and the planet but does it say they are touching before they are displaced?? Or is there a picture??

I believe the situation to be two masses, each on a 2m string, with the stings attached at a common point.
One is drawn to one side so its string makes a 20 degree angle with the vertical, while the other is drawn to the other side until its string makes an angle of 10 degrees with the vertical. After being held there for a moment, the masses are released and they swing back towards each other - ultimately colliding. The question was, what angle with the vertical do the strings make when the masses collide.
 
  • #20
PeterO
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Okay, well we agree on one thing, the masses will collide when they are at 0 degrees. But I have no idea what you're talking about with circular motion equivalent. Mass on spring has the same SHM as a pendulum (at least a pendulum under small angle approximation) and it doesn't move in a circle. The general solution to a second order differential equation x''+x=0 is x(t) or Ø(t) = A*cos(2*π*f+Phase). What I said about the phases for this problem is true, if you don't believe me solve for yourself and get the numerical answer to the transcendental.

Edit: Oh, I think maybe I see your point. As netgypsy says, there was no explicit reference of the angles to a vertical or anything, I was assuming the angles were with respect to the vertical.

One of the simplest introductions to simple harmonic motion was a film of a small sphere undergoing SHM, from side to side.

The camera is then swung around through a 90 degree angle so as to look at the small sphere stuck to the turntable of a record player.

The sphere is undergoing circular motion, and a regular angular velocity of ω radians per second; but when looked at from the side - it undergoes what we call simple harmonic motion.

The equations of SHM can thus be derived as a component of the circular motion. I t was that circular motion comparison I was refering to.
 
  • #21
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Thanks. I figured as much but it wasn't stated and didn't see a picture and knowing some of the weird problems I've dreamed up I never take anything for granted.

Seems like 20 degrees is a bit large for the simple assumption to hold.
 
  • #22
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Thanks. I figured as much but it wasn't stated and didn't see a picture and knowing some of the weird problems I've dreamed up I never take anything for granted.

Seems like 20 degrees is a bit large for the simple assumption to hold.

Yeah, I mean

sinØ~Ø-Ø^3/6+...

So there's an extra 2% that's left out. The small angle approximation won't work so well for 20º. If I get really motivated maybe I'll Runge-Kutta a solution.
 
  • #23
1,506
18
I agree with PeterO and would emphasise the value of the associated circular motion that can always be used to help visualise SHM.
The pendulum at 20 will have an amplitude ( displacement) of 0.684m and the pendulum at 10 will have an amplitude of 0.347m. On a rotating disc representing the SHM these will be 2 diametrically opposite points and they will cross the vertical diameter of the circle together
 
  • #24
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I agree that the small angle approximation introduces an error of about 2%
This would need calculations to 3 significant figures to be seen which I would take to be too small to worry about
 
  • #25
Matterwave
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I think you guys are way over analyzing this. The two objects are off by a phase of pi, since one starts at maximum and the other starts at the negative maximum. The correct equation should then be exactly what the OP posted. What the OP neglected in his calculation; however, is that if cos(...t)=0 (e.g. when the angle is 0!), he cannot cancel the cosines on both sides. This is another classic divide by 0 "paradox".

There is no transcendental equation to worry about here.
 

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