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Simple pendulum: does the period change in a horizontally accelerating frame

  1. Jan 14, 2008 #1
    [SOLVED] Simple pendulum: does the period change in a horizontally accelerating frame

    1. The problem statement, all variables and given/known data
    Simple pendulum which is 5.0m long. What is its period when it is in a truck accelerating horizontally at 5.0m/s/s?

    2. Relevant equations
    w = sqrt(g/d)
    w = 2pi/T

    3. The attempt at a solution

    I want to say it is simply 2pi/T = sqrt(9.81/5.0m), but I am unsure if that equation applies in a horizontally accelerating frame of reference. Any advice appreciated, thanks!
  2. jcsd
  3. Jan 14, 2008 #2

    Shooting Star

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    What is the "effective" gravity in the truck frame due to non-inertial forces?
  4. Jan 14, 2008 #3
    I'm sorry I have no idea what you're asking. lol

    Doesn't this depend on the orientation of the plane of the pendulum's oscillation to the horizontal acceleration? That's all the question gave us though...
  5. Jan 14, 2008 #4
    The truck's acceleration simply is added to the gravitational acceleration. When you add these vectors, g changes (and thus the period).
  6. Jan 14, 2008 #5
    I assume you mean a = (g^2 + 5^2)?
    If the pendulum is oscillating in a plane perpendicular to that truck acceleration vector, it shouldn't change it at all, shouldn't it? As in, the entire pendulum would be pushed backwards while still swinging with the same period.

    If the pendulum oscillates in a plane non-perpendicular with that truck acceleration vector then I understand.

    If I'm wrong please correct me, thank you.
  7. Jan 14, 2008 #6
    a^2 = (g^2 + 5^2)
    so then you change the magnitude of the apparent g on the pendulum.
    It it's perpendicular... it becomes horribly complex
  8. Jan 14, 2008 #7

    Shooting Star

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    The accelerating truck is a non-inertial frame, and to every real force we have to add pseudo forces. If you let an object fall in the frame of the truck, instead of falling vertically, it will fall diagonally to the rear, since the truck itself is accelerating forward. The overall effect is as if there's a gravity field acting backward whose magnitude is 5 m/s^2, plus the real gravity of earth.

    The net result is a g field of sqrt(9.8^2 + 5^2). The mean position of the pendulum is now pointed diagonally to the rear, at an angle arctan 5/9.8 to the vertical.

    The plane of oscillation is the vertical plane containing the direction of the truck's motion.
  9. Jan 15, 2008 #8
    The problem you have here, is that there is a different acceleration acting here, just as Shooting star said.

    But, in the case of a simple pendulum, you could use the formula:

    T = \frac{2\pi}{\omega}

    which you can't use here because: i. The direction of the net force with that of the radial vector of the bob has changed. This shouldn't be much of a problem, but it is because: In the original equation, you could approximate [itex]sin(\theta) \approx (\theta)[/itex] which you can't do here because the direction depends on the ratio of the acceleration of the frame and the acceleration of gravity. In this particular case of a = 5; g = 9.8, the percentage approximation of [itex]tan(\theta)[/itex] as compared to [itex]\theta[/itex] is around 91.1%. This is too poor an approximation.. something around 99% and above is the minimum %age for an acceptable approximation.

    ii. And since you cannot do the approximation stated above, this pendulum basically does not execute SHM. You'll have to use equations for a mathematical pendulum in order to solve this.
    Last edited: Jan 15, 2008
  10. Jan 15, 2008 #9

    Shooting Star

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    This formula can always be used since it is the definition of omega.

    The pendulum executes as much SHM as it does under ordinary conditions, that is, for small oscillations.

    Nothing changes except g has to be replaced by sqrt(g^2+a^2).
  11. Jan 15, 2008 #10
    Agreed. my mistake though, I was referring to equating the [itex]\omega[/itex] with [itex]\sqrt(\frac{g}{l})[/itex]

    I don't think i'll agree with you on that. Because for the case of a simple pendulum [inertial frame; non-accelerated], the angle made by the force with the mean position was given by [itex]\theta[/itex] i.e. the angular displacement of the length from the mean position. This angle was then used to calculate the torque and after approximation gives an SHM-type motion.

    However, when accelerated, the direction will be given by:

    tan\left(\epsilon + \left(\frac{\pi}{2} - \theta\right)\right) = tan(\gamma)~~~(say)

    tan(\epsilon) = \frac{g}{a}

    where [itex]\gamma[/itex] is the angle between the radial vector of the bob and the net force acting. For the pendulum to be in SHM [or angular SHM], we should be able to prove that the Torque is directly proportional to the angular displacement i.e. [itex]\theta[/itex]. But, torque here is directly proportional to [itex]sin(\gamma)[/itex], which is not directly proportional to [itex]sin(\theta)[/itex] (or [itex]tan(\theta)[/itex] for small [itex]\theta[/itex]).

    With some trignometry, you can get:

    sin(\gamma) = sin(\epsilon)(cot(\epsilon)cos(\theta) + sin(\theta))

    Which means, that [itex]sin(\gamma)[/itex] is proportional to [itex]sin(\theta)[/itex] when [itex]cot(\epsilon) \to 0[/itex] i.e. [itex]tan(\epsilon) \to \infty[/itex] i.e. [itex]\epsilon \to 0[/itex], which basically means a = 0 i.e. the frame is non-accelerated.

    So, when the acceleration of the frame tends to 0, the oscillations show harmonic tendency.
    Last edited: Jan 15, 2008
  12. Jan 15, 2008 #11

    Shooting Star

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    The whole frame behaves as if there's a new gravity field g' in the specified direction. All calculations you have done to derive the period of a simple pendulum in a static frame is valid in this frame with the new g'. There's no need to go into all that math.

    Read up on non-inertial frames, if you are not satisfied.
    Last edited: Jan 15, 2008
  13. Jan 15, 2008 #12
    Thanks for the help guys. I asked my prof today at the end of lecture (not to doubt you!) and he said the same thing, helped me to understand why no matter what the orientation the additional velocity vector will simply be added to the gravitational force vector. i.e. a^2 = (g^2 + 5^2).

    Thanks shooting star, rohanprabhu.

    edit: oops thanks for adding in [solved], mod.
    Last edited: Jan 15, 2008
  14. Jan 16, 2008 #13
    i asked my physics lecturer too.. and well.. even he agrees with you. He says that, the pendulum will still act as an harmonic oscillator, but the mean position around which it oscillates will be different. But, the same formula can be used and you just have to substitute the new 'g'.

    so, i swallow my pride and congratulate you :approve:
  15. Jan 16, 2008 #14

    Shooting Star

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    Found this funny quote:

    Swallow your pride occasionally, it's non-fattening! ~Author Unknown

    Didn't somebody say that humility is the way to greatness? You'll go places, rohanprabhu.

    Best wishes.
  16. Jan 16, 2008 #15
    thx.. btw.. i'm just posting this message to get to my 100th post :D
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