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4points Cerjonic moticn' simpif pendulum oscillates iith amplitude andthe frequency f1. If the length of the pe nquUM/S dcubled whatisthe neiv frequencyofits m...

Question

4points Cerjonic moticn' simpif pendulum oscillates iith amplitude andthe frequency f1. If the length of the pe nquUM/S dcubled whatisthe neiv frequencyofits motion? fo}v22 Ja vfo fo/ 2

4points Cerjonic moticn' simpif pendulum oscillates iith amplitude andthe frequency f1. If the length of the pe nquUM/S dcubled whatisthe neiv frequencyofits motion? fo}v2 2 Ja vfo fo/ 2



Answers

A simple pendulum oscillates with frequency $f .$ What is its frequency if it accelerates at 0.50$g(a)$ upward, and (b) downward?

For this problem on the topic of oscillators, we are told that a simple pendulum is oscillating with a frequency of F. We want to calculate the new frequency if the entire pendulum accelerates with half the acceleration due to gravity firstly upward and then downward. Now the frequency of a simple pendulum wino is given by F. This is one over two pi times the square root of G over L. So the pendulum is accelerating vertically, which is equivalent to increasing or decreasing the acceleration due to gravity by the acceleration of the pendulum. So let's first look at the scenario in which the pendulum is accelerating upward, so we'll call the new frequency F. New, This frequency is one over two pi. I'm the square root of G plus A. Of our, So this is one over two pi times the square root Of 1.5 G. Over our, or simply The square root of 1.5 times one over two pi times the square root of G over L. So the term in brackets is simply F. So the new frequency is the square root of 1.5 times the old frequency F. Which is simply 1.22 times the original frequency f. So the new frequency is greater than the old frequency. Now, what happens if the pendulum is accelerating downward? So some little above the new frequency F. New is one over two pi times the square root of G plus A. Over L. Hey in this case is going to change, that's one of the two pi times The Square Root of G -0.5 G. Which gives us 0.5G, divided by L. And again we can write this as the square root of a half Skirt of 0.5 Times 1/2 Pi. And the square root of G. Of L. Of the time in brackets again, is the original frequency F. This becomes a square root of a half times the original frequency F. And so this becomes 0.71 times F, which is now less than the original frequency.

So remember, if we're given an equation in the form of of a co sign of B T, then our amplitude is going to be our value of a our period is going to be two pi over B and then our frequency is going to be the reciprocal of our period. So we're gonna have be over two pi and so are amplitude in this. This equation is force or amplitude is four. Then we have our period. This is two pi over pi over two and two pi over. Pi over two gives the value of of four. So our amplitude is for our period is for and our frequency is going to be the reciprocal of our period. So this is going to be one over for hurts. So these are solutions.

So remember, if we're given an equation in the form of a sign of B t, then our amplitude is going to be Our amplitude is going to be a and R period is going to be two pi over beat and our frequency is going to be the reciprocal of our period. So this is going to be be over two pi. So are amplitude in this in this expression, is is 11 we have amplitude is 11 are period is going to be two pi over 12 pie. So two pi over 12 pie gives us to over 12 which is equal to 1/6. And so since our frequency is the reciprocal of our period than our frequency is going to be six hertz and so this is our solution.

So we're gonna set equation 13 6 equal to the maximum velocity divided by two. So we can say that the maximum velocity multiplied by sine of two pi multiplied by the time t divided by capital t the period This is gonna be equal to the maximum velocity divided by two. And so we can then say that sign of to pie t over capital T the period equals 1/2. Now we can then sulfur t where t would be equal to the period divided by two pi multiplied by arc sine of 1/2. This is gonna be equal. Two tea over to pie ark Side of 1/2 is pi over six, about 30 degrees pi over six. The pies were gonna cancel out and we have the period over 12 or because of the period this period is city T over 12 or five t over 12. So we can say that subtracting the first time from the second time and then multiplying my too. We can simply say two times five t over 12 minus t over 12 is gonna be equaling eight t over 12 or to t over three so we can say that the odds are the the object speed is greater, then the maximum velocity divided by two core 2/3 of one sec. That is the end of the solution. Thank you for watching.


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