A)simple pendulum is displaced (+8 degrees) away from its vertical (equilibrium

Question

A)simple pendulum is displaced (+8 degrees) away from its vertical (equilibrium ) position and is held at rest in that position. at t=0, it is released, whereupon it proceeds to oscilate about the vertical with freqency of 2Hz. What is the position of the pendulum( its angle with respect to the vertical ) at A) t=1.3s, B)t=13.0s?. consider angle to be positive when the pendulum is to the right of the vertical. consider angle to be positve when the pendulem to the left of the vertical.

B) for the same problem above, what percentage of the time is the pendulum more than (5 degrees ) away from the vertical? that is, what percentage of the time [theata] is bigger than 5 degrees. Hint, the easiest route is probably the one based on the concept of simple harmonic motion as the projection of uniform circular motion onto diameter of circule.

Explanation / Answer

(A)The amplitude of the oscilation is 8degrees θ=Acos(wt-B) is the general equation for shm where A=8(amplitude) B=phase difference w=frequency as at t=0 it is released from rest B=0 and f=2Hz,f=w/2pi which means w=4pi rad/s so at at t=1.3s θ=8cos(4pi*1.3)

so θ=7.67 degrees

When t=13 s θ=8cos(4pi*13) so θ=-7.66(to the left) degrees

when left is positive and right is negative it will be just opposite to the above case

(B)We consider the concept of simple harmonic motion as the projection of uniform circular motion onto diameter of circule so for it to be more than 5 degrees the θ which the object moving in the circular path makes with the horizontal is 5=8cos(θ), θ=51 degrees so from 51 to 90 degrees it will be above 5 degees so time is (90-51)/w and it happens twice so time is 2*(90-51)/w which is 78/w and total time is 360/w so the percentage of time for which it is above 5 degrees is 21.66%

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