2. [15 marks]: Longitude - an exercise in non-inertial reference frames. In class we discussed how to find your latitude

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2 15 Marks Longitude An Exercise In Non Inertial Reference Frames In Class We Discussed How To Find Your Latitude 1 (123.25 KiB) Viewed 11 times

2. [15 marks]: Longitude - an exercise in non-inertial reference frames. In class we discussed how to find your latitude, should you find yourself enclosed in a box with a bunch of exper- imental equipment, using the Coriolis force. A historically more profound measurement is longitude. Unlike latitude, the fictitious forces from non-inertial frames are the problem, not the solution. This was first solved in the 18th century by the development of "good" clocks (see, e.g., Longitude by Dava Sobel). Here we will explore why this is so hard! The picture you should have in your head is a pendulum (our "clock") on a table that shakes side-to-side (our simplified "boat"), i.e., X e 8 Xp
Xb The pendulum has length and mass m, and has a swing angle 8 from the vertical. The complication is that the entire system is accelerating side-to-side as the boat pitches about on the waves (we are ignoring the rolling and up-down motion for simplicity). a. (2 marks) Write down expressions for the pendulum position in an the inertial frame in terms of an arbitrary left-right shift as a function of time. b. (3 marks) Show that a resulting Lagrangian for the pendulum is L= m[²6² +2isl cos(0)] + mgl cos 0. Note that some terms have been dropped from the Lagrangian in the inertial frame (i.e., L = (m/2)||2mgy), note which ones and identify why. TU c. (2 marks) Write down the equation of motion for 8 (ie., the 2nd order differential equation for 6). Show that in the small-angle approximation, i.e., 0 < ^, 1 cl. 04
for 6). Show that in the small-angle approximation, Le., U. 17 0 + 20 = l This is the equation of motion of a forced harmonic oscillator. d. (4 marks) The ship is constantly buffeted by waves, wind, storms, etc., and thus does not executed nice, ordered motions. Nevertheless, we can expand the boat motion into a set of harmonic components: Tb (t)=[b, sin (2 t) + cj cos(12,t)]. Show that in this case the following is a solution for 0, assuming the pendulum is started at t = 0 at height 0 if w, BJ, and C, are chosen correctly (i.e., find w, B,, and C₂). 0(t) = 0, cos(wt) + Σ [B, sin(t) + C, cos(,t)]. e. (4 marks) Pendulum clocks work because there is a known one-to-one relationship between and time. However, we now have a bunch of additional junk from the motion of the boat. Let us define Otrue (t) = 0o cos(wt). Show that the time-averaged shift in the mean of 0 is zero, i.e., but the root-mean-square error is not, i.e., ((0-0)²) 0. For this you may assume that (sin(t)) = (cos(t)) = 0 and (sin('t)) = (cos('t)) = 1/2 for all (not to be confused with ). In particular, evaluate the RMS error in terms of the b,/, c/w, and St. The solution is to not use pendula, of course!