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G, T of pendulum

  1. Aug 17, 2004 #1
    There are a couple of problems from my text, each focusing around g and T for a pendulum, that I feel unsure about my answers, especially since each is listed as a "challenge problem". The questions are short so I'll just rewrite them right here.

    1) A simple pendulum clock keeps correct time at a point where g = 9.8000 m/s², but is found to lose 4.0 s each day at a higher elevation. Use the result dT/T = (-1/2)(dg/g) to find the approximate value of g at this new location.

    ans. 9.7791 m/s²

    2) You measure the period of a physical pendulum about one pivot point to be T. Then you find another pivot point on the opposite side of the center of mass that gives the same period. The two points are seperated by a distance L. Use the parallel-axis theorem to show that g = L(2pi/T)². (This result shows that you can measure g without knowing the mass or moments of inertia of the physical pendulum.)

    A1) Allright, I simply plugged the numbers in and didn't get nearly quite that large a difference in g. Thinking about it more carefully, let's suppose this simple pendulum has a period T = 2.00 s where g = 9.8000 m/s². Each day, or every 86400 s later, it thinks only 86396 s have passed. If it thinks 86396 s have passed, then it must have counted 43198 cycles. Then we have that 86400 s have passed in the course of 43198 cycles. From this we determine dT = 0.00009260. Okay, now that I feel pretty confident dT is correct, I plug into the formula: dg = -2(dT/T)g = -0.0009074 m/s² or g + dg = 9.7991 m/s².
    Not jiving with the book.

    A2) Okay, after frying a few neurons in search of a proof, I thought I might look for a counterexample. After a couple of trys I came up with this: take a uniform ball (solid sphere) of diameter L as the physical pendulum and let the pivot points be any two points whose connecting line forms a diameter. Note that the pivot points satisfy all the hypothesis of the problem. Now the period T = 2pi sqrt( I/mgx ) for small amplitude oscillations and where I is the inertia about the pivot point and x is the distance of the center of mass to the pivot point. Rearranging we find g = (2pi/T)² (I/mx). In this case g = (2pi/T)²[ (2/5m(L/2)² + m(L/2)²]/[m L/2] = (2pi/T)²[7L/10]. This does not equal what the book suggests, (2pi/T)² L.
  2. jcsd
  3. Aug 17, 2004 #2


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    No, dT is an infinitesmal. Using "dT= 0.00009260" gives an approximation (and not a very good one). To use dT/T= -1/2 dg/g, integrate both sides to get ln(T)= -1/2 ln(g)+ C or T= cg-1/2. Writing the two periods as T1 and T2 (at the higher altitude) we can get rid of the c by dividing T1/T2= (g1/g2)-1/2 or g2/9.8= (T1/T2)2. You could take T= 2 sec but, since T does not really matter and we are given that it "loses 4 sec every day" I think it is simpler to take 1 day or 86400 s as "T1" and 86396 s as "T2". That gives g2/9.8= 86400/86396. g2= (86400/86396)(9.8)= 9.80045 to 5 decimal places.

    A counter example?? Do you have any reason to think that the given result is not true? You might start by writing out precisely what the "parallel axis theorem" says.
  4. Aug 17, 2004 #3
    I don't think your first result is correct and I also think the approximation works very well because the change in the period is minute. As a simple check, if we are going to a higher elevation then we expect g to decrease. The contradiction arises from an improper assignment of T2. If T2 were 86396 s, then the clock will count a full day as passing four seconds before it does pass (i.e. gain time). I think the correct T2 in your case is T2 = (86400 s)²/(86396 s). If I ignore the approximate formula and use the more precise relation g2 = (T1/T2)² g1 with T2 = 2.00009260 s and T1 = 2.00000000 s, then I get virtually the same result as before. I'm beginning to think that the answer is just mistyped by a digit.

    As for the second one.. yeah, i don't know, it kind of throws it in the face of the problem to say that the statement is not true. But I've been pretty unsuccessful in trying to prove it and trying to disprove it seemed to be rather easy. At first I thought maybe I am misinterpreting the problem, but I read it over carefully, and I just don't see much room for ambiguity. If what the book says is correct, then someone should be able to point out an error in my counterexample.
    Last edited by a moderator: Aug 17, 2004
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