This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: Quiz 1 Solutions Nate Bode October 31, 2009 Note: The first thing we should do is take a look at the Hookes Law equation that the torsional springs obey. The first thing we should note is that [ i ] = Force Length , so we know we are dealing with a situation unlike those seen in class and in sections (though an example was given during the quiz review). Therefore, because 1 and 2 are unitless, we should also take note that [ ] = Force Length . The only other physical constants given are g which has units of acceleration and the mass and length of the pendulum L which obviously have units of mass and length, respectively. The only way we can get a frequency out of these is by the combinations p g/L and p /mL 2 . The symmetric mode has both masses swinging in phase which means that the spring is doing nothing. Therefore we know that symmetric = p g/L. The antisymmetric mode is a little more complicated, but if you think carefully you will know that this problem is now completely analogous to two pendula coupled by a standard spring, a problem we have done before which by direct analogy k/m /mL 2 gives us the antisymmetric modes frequency: antisymmetric = r g L + 2 I , where I = mL 2...
View Full Document