The quantities A and p (called the amplitude and the phase) are undetermined by the differentialequation. They are determined by initial conditions -- specifically, the initial position and the initialvelocity -- 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 noposition or velocity. Consequently, A and p (and also yo) are irrelevant in the problem. We onlycompare the period T or the frequency w with the theoretical prediction. You have (hopefully)derived (or maybe looked up) the relation between w and k and m.This final question relates w and T. If w = 5.8*10° rad/s, calculate T in seconds. (Remember, that aradian equals one.) T might be a fraction of a second.

Given the angular velocity of the pendulum: ω=5.8×100 rad/s=5.8 rad/s

Answered: The quantities A and p (called the…

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A meter stick of total length l is pivoted a distance d from one end on a friction-less bearing. The stick is suspended so that it becomes a pendulum. This is called a "physical" pendulum because the mass is distributed over the body of the stick. Assume the total mass is m and the mass density of the stick is uniform. The acceleration of gravity is g. Find T and V as functions of generalized coordinate θ and velocity ˙θ. Do this by considering the stick to be divided into infinitesimal parts of length dl and integrating to find the total kinetic and potential energy. Set up the Lagrangian and find the equation of motion.
part A and B
For Parts I and II, use the pendulum equation and solve to predict the length needed for a pendulum with a period of 1 s and 2 s respectively. Use a piece of string (or thread, dental floss, shoe lace, etc.) and a steel nut (or washer or something small but with enough mass to weigh down the string) to build each pendulum. Time each pendulum for 30 periods and then find the average time for one period. Remember, a period is the time to complete one cycle of motion, out and back. Tape each pendulum up in perhaps a doorway where it is stationary and has room to swing. Be precise with your measuring and timing. Show your work. Part I Given: T = 1.00 s l =? Part II Given: T = 2.00 s l =?
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.
An electron undergoes simple harmonic motion with the acceleration shown below: ax(t)=−amaxsin(2t/T) with amax=5839 ms2 and T=316 seconds. Assuming that the only motion is oscillatory (ignoring overall translation), what is the maximum speed of the electron? What is the amplitude of the electron's position?
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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