A pendulum is made by tying a 75 g ball to a 130-cm-long string. The ball is pulled 5.0° to the side and released. How many times does the ball pass through the lowest point of its arc in 7.5 s?

Video transcript

Welcome back, everyone. We are making observations about a Bob and we are told that it has a mass of 125 g or 1250.125 kg kilograms, 125 kg. Now, we are also told that it's suspended from a massless thread, 100 and five centimeters long, 41.5 m long. Now, we have an initial displacement of eight degrees and we are tasked with finding after a time of 10 seconds, how many times will the Bob pass through the e equal Librium position? Now, let's think about this logically here, right? We need some way to relate these variables to the number of times it passes through the equilibrium. Well, if we're starting from an initial displacement off the equilibrium, that means as soon as we drop it at a point in time, that is 1/4 of the period, the ball will pass through this center equilibrium point, then it'll pass through again at three T over four, then five T over four and so on what this gives us is that at any time T we can say that T is equal to one plus two N times the period divided by four or N equals zero, comma one, comma two and so on. So this relates the period to the number of times it passes through the equilibrium. But what is our period? Well, we know that angular velocity can be given by two pi over the period. And we also know that angular velocity is equal to gravitational acceleration divided by the length of the chord. So using the right two equations and solving for T what we get that is that T is equal to two pi times the square root of the length of the chord divided by our gravitational acceleration. Now plugging that into our equation for time, we have one plus two, N times two pi over four times the square root of L over G solving for N what we get is T times the square root of G over L times 4/2 pi minus one all divided by two great. So now we have finally an equation for the number of times it will pass through the equilibrium. Let's go ahead and plug in T here. All right. So we have N is equal to 10 times the square root of 9.8 divided by one point oh five times or over two pi minus one all divided by two, plugging this in, we get 9.22, which means N is equal to zero, comma one comma two all the way up to nine. And if you count up the number of NS, what we get is that N is equal to 10, which corresponds to our final answer. Choice of a. Thank you all so much for watching. I hope this video helps. We will see you all in the next one.