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.