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Derive an equation

  1. Mar 24, 2014 #1
    I need some help with an equation.

    I will use to find the value of acceleration due to gravity.

    With a air track and a spring attached to a glider we should find the value of "g".
    The track is inclined and with two different equilibria (which are achieved by using two different masses on the glider) we kan determined "g".

    g=(4pi[itex]^{2}[/itex][itex]\cdot[/itex]bL)/((T[itex]^{2}[/itex]-t[itex]^{2}[/itex])H)

    b = the extension of equilibrium (we got it to 0.026 meters)
    L = the length of the track (2 meters)
    T= the time of one period with more weights on the glider (2.37 s)
    t = the time one period with no extra weights on the glider (2.12 s)
    H = the height of the air track (0.184 meters)


    If we insert all the values we get 9.939m/s/s.

    How should we derive the equation?
    If we take the formula of pendulum motion we get g=(4pi[itex]^{2}[/itex][itex]\cdot[/itex]L)/T[itex]^{2}[/itex]

    I don't seem to get any further with the equation so I need some help.
     
  2. jcsd
  3. Mar 25, 2014 #2

    Philip Wood

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    Gold Member

    This is quite tricky if you've just started to study SHM.

    The key thing to understand is that you're not dealing with pendulum oscillations but with oscillations in a straight line of a mass on a spring.

    So the key equations are [itex]T=2\pi \sqrt \frac{m+m_0}{k}[/itex] and [itex]T_0=2\pi \sqrt \frac{m_0}{k}[/itex].
    The first of these is with the extra mass; the second is without. If you square these equations and subtract them you'll get something useful which doesn't involve [itex]m_0[/itex].

    When you add mass to the glider the oscillations are at a lower rate, because of the inertia of the mass, not because of the extra gravitational pull on the mass. But the gravitational pull does make the glider + mass equilibrium point lower down the slope than with glider alone. If it sits a distance b lower down the slope (so the spring is extended by a further distance b) you can write a 'static' equation which links k, b, m, g, H/L. This enables you to eliminate k and m from your oscillation equation.

    Good luck!
     
    Last edited: Mar 25, 2014
  4. Mar 25, 2014 #3
    Ok, so I get:

    k=(4pi^2*m) / (T^2-t^2)

    and because Hooke's law says k = F/X (I'll use b as notation instead of X) F=mg so k=mg/b

    g = (4pi^2*b) / (T^2-t^2)

    How do I get the (H/L)-term into the equation?
    Have it something with Sin(x)=H/L so 1/Sin(x)=L/H to do?
     
  5. Mar 26, 2014 #4

    Philip Wood

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    Gold Member

    Yes.

    No. mg is not the force stretching the spring (at equilibrium) because the system is on a slope. The effective force stretching the spring is mg sin(θ) = mg (h/L). [Actually this is the EXTRA force due to mass m being added to the glider and b is the extra distance stretched by the spring.]

    Yes. See above.
     
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