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Time period of system with constant restoring force

  1. Oct 31, 2016 #1
    1. The problem statement, all variables and given/known data
    Two smooth planes are joined at one end so that they form a V shape. The join is such that a mass placed on one of the planes will slide smoothly down one side of the V and then move up the other side. Find the period of the motion (T) of such a mass in terms of x0 (the initial horizontal position) and θ (the angle between each of the planes and the horizontal.

    2. Relevant equations
    The potential energy in terms of horizontal position and hence the force and acceleration.

    3. The attempt at a solution
    The potential energy of the mass on the LHS of the well seems to go like...
    V(x) = m.g.(x0-x).tan(θ) where x is the horizontal displacement from x0. I set things up like this to give V(x) = 0 when the mass is at the lowest point.
    and by differentiating I get a horizontal force m.g.tan(θ) towards the centre and an acceleration of g.tan(θ).

    Since the mass is uniformly accelerated through horizontal distance x0 in a time of T/4 I write
    x0 = 1/2 . g.tan(θ) . (T/4)2 (assuming the initial speed is zero)
    and find that
    T = √[32x0/g.tan(θ)] = 4√[2x0 / g.tan(θ)]

    I would like to know if this is correct. It seems like it should be the sort of thing you could just look up, but I don't seem to be able to describe the system well enough to Google to get results about anything other than harmonic oscillators.
     
  2. jcsd
  3. Oct 31, 2016 #2
    I did not get the same answer when I worked the problem in, what is probably, the more standard method. Hopefully one of the staff guys will chime in.
     
  4. Nov 1, 2016 #3

    haruspex

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    So if θ>π/4, the acceleration would exceed g?
     
  5. Nov 1, 2016 #4
    Yeah, I was thinking about this last night and my answer doesn't make any sense. The 1/tan term drives T to zero as the angle goes to π/2 (which as you say, is because the acceleration goes off to ∞ with increasing θ). Clearly the tan term is wrong.

    I tried this a different way by just using the component of the weight force down the slope and I ended up with sin(θ).cos(θ) instead of the tan(θ) term in the final result. This seems more likely as sin(θ).cos(θ) goes to 1 rather than ∞.

    I shall go back and take a look at my original thinking about the potential and post the result later on this evening.
     
  6. Nov 1, 2016 #5
    ...except sin.cos doesn't go to 1 and I am a first rate fool for posting such nonsense. It goes to 0.5 at π/4 and this would have the effect of minimising T at this angle.

    I'm going to start again. Don't tell me, I'll get it in the end!
     
    Last edited: Nov 1, 2016
  7. Nov 1, 2016 #6
    Okay, here's what I've got. I pondered for a long time and felt like I could feel the shape of the answer but couldn't see it. In fact, I wrote all this as I was thinking, which is why this post is so long and rambling.

    As the particle slides down the slope, work is done by gravity. For each change in horizontal position δx the displacement of the particle is δs = δx/cos(θ) and the work done by gravity is mg.sin(θ) δs = mg.sin(θ) δx / cos(θ) = mg tan(θ) δx. Clearly this only makes sense if θ<π/2. For θ = π/2, no horizontal motion is possible. For large θ, δx gets small (for fixed starting height, h0). In fact, if the mass is always released from the same height above the lowest point in the track then the initial horizontal displacement (x0) is h0/tan(θ). So the total work done by gravity as the mass slides down the slope is always mg tan(θ) . h/tan(θ) = mgh. This is what you'd expect.

    It seems easier to set zero potential energy at the lowest point. If we move away from this point we do work against gravity like mg.sin(θ) δs = mg.tan(θ).δx which means that at any position x the energy is mgx.tan(θ). Differentiating this gives a result for the restoring force that is plainly wrong!

    The problem with this, I think, is that x is not the position of the particle. When we differentiate the potential energy function we need to do so with respect to position - not just one component of the position.

    So using mg.sin(θ).δs looks to be a better option because i)I can see that it's going to give a better answer and ii) δs is the actual change in position.
    dV/ds = mg.sin(θ) = -F = -ma which certainly looks right as θ→π. This is an elementary result and of course we can now solve the uniform accelerated motion to get

    T = 4√[2s0/g.sin(θ)] = 4√[2x0/g.cos(θ).sin(θ)] which is (of course) exactly what you get if you just resolve the force down the slope in the first place! (Imagine a particle in space accelerating due to a constant force = mg.sin(θ) etc.)

    For a fixed x0, T has a minimum at θ = π/4. Why? Well, for small θ the displacement is minimised towards x0 but the force goes towards zero. For very large angles the force goes towards mg but the displacement runs off to infinity. Beyond π/4, the force increases more slowly but the displacement increases much faster.

    Now somebody is going to come along and tell me that I'm wrong.
     
  8. Nov 2, 2016 #7

    haruspex

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    Not me.
     
  9. Nov 2, 2016 #8
    Then I am content!
     
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