1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

At what angle will two pendulums collide?

  1. Jan 27, 2012 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

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

    3. 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,
     
  2. jcsd
  3. Jan 27, 2012 #2
    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.
     
  4. Jan 27, 2012 #3

    PeterO

    User Avatar
    Homework Helper

    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?
     
  5. Jan 27, 2012 #4
    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.
     
  6. Jan 27, 2012 #5

    PeterO

    User Avatar
    Homework Helper

    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?
     
  7. Jan 27, 2012 #6
    That they will hit exactly between the two starting positions?
     
  8. Jan 27, 2012 #7

    PeterO

    User Avatar
    Homework Helper

    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.
     
  9. Jan 27, 2012 #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...
     
  10. Jan 27, 2012 #9

    PeterO

    User Avatar
    Homework Helper

    I hope you saw my edit.
     
  11. Jan 27, 2012 #10
    Well exactly between both starting positions is +5° right?
     
  12. Jan 27, 2012 #11

    PeterO

    User Avatar
    Homework Helper

    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.
     
  13. Jan 27, 2012 #12

    PeterO

    User Avatar
    Homework Helper

    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.
     
  14. Jan 27, 2012 #13
    I understand... Thank you very much
     
  15. Jan 27, 2012 #14
    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.
     
  16. Jan 27, 2012 #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...
     
  17. Jan 27, 2012 #16

    PeterO

    User Avatar
    Homework Helper

    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.
     
    Last edited: Jan 27, 2012
  18. Jan 27, 2012 #17
    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??
     
  19. Jan 27, 2012 #18
    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.
     
    Last edited: Jan 27, 2012
  20. Jan 27, 2012 #19

    PeterO

    User Avatar
    Homework Helper

    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.
     
  21. Jan 27, 2012 #20

    PeterO

    User Avatar
    Homework Helper

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook