Simple Harmonic Motion with 2 beads

  • #1

Homework Statement


A small bead of mass m1 slides without friction in a spherical bowl of radius R. The bead is displaced a height h1 from the bottom of the bowl. A second, more massive bead of mass m2 is displaced a height h2 from the bottom of the bowl opposite the first bead. The two beads are released at the same time.

Calculate the period of the harmonic motion, individually, for each of the beads as if they are placed alone in the bowl.

Identify the position where the two beads first meet.


Homework Equations


T = W / 2pi
x = A cos/sin (Wt + delta)


The Attempt at a Solution


My professor mentioned using the formula for the angular frequency of a mass spring system, that being the square root of k / m. However, I do not see its applicability at this point. I have no idea how to calculate the period with the information given...for instance, how am I to determine the angular velocity based upon what is given? I figured that was the place to start, but again, I don't see how the sq. root of k / m will help me since we're not dealing with a spring. As for the second part, my best guess is that I need to write position equations for each bead and set them equal to one another? Is that right?

Please help! :bugeye:
 

Answers and Replies

  • #2
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A good place to start would be with conservation of energy.
 
  • #3
HallsofIvy
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You could also calculate the force acting on the bead along the tangent to the bowl. If [itex]\theta[/itex] is the angle the radius from the center of the bowl to the bead makes with the horizontal, then the force parallel to the bowl is [itex]-mg cos(\theta)[/itex]. Setting that equal to ma will give you an analog to the spring problem. However, a spring, acting horizontally as is normally the case in "Simple Harmonic Motion" problems, does not involve gravity. This does and, as always in gravity problems, the mass cancels out- the two beads will accelerate at the same rate. If, for example, h1 and h2 were the same, the beads would collide at the bottom of the bowl.
 
  • #4
nrqed
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A good place to start would be with conservation of energy.

As far as I can see, conservation of energy is useless for this particular question. The beads won't have the same speed when they collide.

HallsofIvy gave the only way to proceed: using Newton's second law.
However, I woul duse the angle defined between a line going from the center of curvature of the bead and the verticla, in which case the force will be [itex] - m g sin \theta [/itex]. Assuming small angles, the sin can be approximated by the angle itself. The next step is to write the acceleration as [itex] r \frac{d^2 \theta}{dt^2} [/itex]. This way, one gets an equation which is similar to the one for a mass attached to a spring. One may then directly obtain (by analogy) the equation for the angle as function fo time [itex] \theta(t) [/itex]. As HallsofIvy pointed out, the mass will have cancelled out in the equation and one finds out that the angular frequency [itex] \omega [/itex] is the same for both beads. The only difference will be the angular amplitude of oscillation, [itex] \theta_{max}[/itex]. These can be found using the initial heights and simple trigonometry.

But the fact thatteh two angular frequencies are the same should suggest something interesting that will happen!
 
  • #5
I will definitely try the force approach. I can't believe I didn't think of it. I guess I just panicked when I saw the problem. Thanks for all of the input!
 
  • #6
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I suppose that short of a Legrangian conservation of energy isn't as helpful as force.
 

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