I have no problem with the solution provided but, I have a problem with understanding it's meaning. Shouldn't the two solutions for b) add up to be the period? If not, why?
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2 Answers
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No. The period is the total time, or if you prefer, the $\Delta t$ required to complete one full cycle. Part (b) is asking for particular points in time at which the particle is passing the equilibrium point.
answered Sep 6, 2020 at 20:21
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It would be. But in this case the equilibrium point and the initial point are not the same.
If we analyse the motion, we see that first the body moves to the right extreme, comes back to mean position, and then moves to the left extreme and the again moves to the mean position along the positive X axis. The time taken in this process is the first answer.
As for the second motion, the body only moves to the right extreme and comes back to the mean position. The time taken in this process is the second answer.
Adding the time taken in these cycles will not give the time period. You will see how some of the motion has been repeated in these cycles. To get the time period we must start and finish at the mean position, making sure that both the extremes are covered.
So, to answer your question: No. The sum of the times will not be the time period in this case because the equilibrium point and the initial point are not the same.
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$\begingroup$ thank you! this clears things a bit although, I need to do some research as I feel I don't quite understand it yet $\endgroup$– AhmedSep 7, 2020 at 13:36