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The generalized coordinates of a simple pendulum are the angular displacement and the
angular momentum is m/26. Find the trajectory in phase space of the system and show that
the area 4 enclosed by a trajectory is equal to product of total energy E and time period of
the pendulum.

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Consider the solution tothe harmonic oscillator given above by x(t)=Ccos(wt−v) Prove tha tx(t0)=x(t0+2piw) In other words, the solution has the same value at time:t0 and at time:t0+2piw regardless of what value we have for ?0. The value 2piw is then the period T of the harmonic oscillator.
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The quantities A and φ (called the amplitude and the phase) are undetermined by the differential equation. They are determined by initial conditions -- specifically, the initial position and the initial velocity -- usually at t = 0, but sometimes at another time. In the oscillating part of the experiment, I measured only the time of 30 periods. I measured no position or velocity. Consequently, A and φ (and also y0) are irrelevant in the problem. We only compare the period T or the frequency ω with the theoretical prediction. You have (hopefully) derived (or maybe looked up) the relation between ω and k and m. This final question relates ω and T. If ω = 8.2*102 rad/s, calculate T in seconds. (Remember, that a radian equals one.) T might be a fraction of a second.
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