(II) Consider two objects, A and B, both undergoing SHM, but w...

Question

(II) Consider two objects, A and B, both undergoing SHM, but with different frequencies, as described by the equations 𝓍ₐ = (2.0 m) sin (4.0 t) and 𝓍₆ = (5.0 m) sin (3.0 t) , where t is in seconds. After t = 0, find the next three times t at which both objects simultaneously pass through the origin.

Accepted Answer

Welcome back. Everyone in this problem. A group of students are experimenting with two different pendulums, pendulum, one and two placed side by side. They set both of the pendulums in swing by pushing them from the equilibrium in the same direction. At the same time, they find that the equations giving the distance of the Bobs from the equilibrium position are X, one equals 1.5 m multiplied by the sine of five T and X two equals 2.5 m multiplied by the sine of eight T. Determine the next two consecutive times after the pendulums begin swinging apart from the first time corresponding to T equals zero at which the Bobs pass the equilibrium positions simultaneously. For our answer choices. A says it's at one second and two seconds. B at 3.1 seconds and 6.3 seconds C at five seconds and eight seconds and D at four, multiplied by 10 seconds and eight multiplied by 10 seconds. Now, before we start, let's just make sure we understand what exactly we're looking for. OK. So we're saying that we have two pendulums that are swinging. OK? So let's call this pendulum pendulum one and pendulum two. Yeah. And for our pendulums, we want to determine after we've pushed them from the equilibrium position, then we want to know when again the pendulums are going to pass the equilibrium position simultaneously. In other words, at our equilibrium positions here, OK. If we let P one be the blue pendulum, OK? And P two be the red pendulum, we want to find when again on their swing, both of them are going to come back to equilibrium based on our diagram here. OK. So that's basically what we're trying to figure out. What time is that going to happen. Now, we already have the equations of motion for our two pendulums. And when the Bobs pass the equilibrium, then the distances X one and X two are zero. Now this will only happen if the argument of the sine functions are multiples of pi. In other words, at X one and X two being equal to zero. OK. Then that means the argument for X one that is the argument for the sine function of X 15 T is going to be equal to a multiple of pi. In other words, we can call that N one pi and, and let's call this equation one here. And the argument for X two which is eight T is also going to be equal to a multiple of pi, let's call that N two pi, OK. Where N one and N two are positive integers. Now based on both equations here, you can tell that from our first equation T is going to be equal to N one pi divided by five. And in our second equation T is going to be equal to N two pi divided by eight. No, these two equations derived from equation one and two give the times at which the pendulums, one and two pass the equilibrium when they pass simultaneously, these times are equal. That means both of these equations are equal. And thus at equilibrium. OK. Let me don't tell anybody. I didn't spell that properly. OK? But here at equilibrium, OK. Then N 15 or rather let me express N two in terms of N one, N two pi divided by eight is going to be equal to N one pi divided by five. Now we can cancel pi from both sides here. And if we solve for N two, then N two or second multiple is going to be equal to eight N one divided by five. Now without figuring out these multiples, remember N one and N two are positive integers. So if N two is going to be equal to eight multiplied by N one divided by five, which we can rewrite this way, then this shows us that N one must be a multiple of five for us to get a corresponding value of N two. Both of these values would yield the same time when plugged into our equations at which both of the pendulums are at equilibrium. So if we are going to find those times, we can take two consecutive multiples of five as N one and find the corresponding times because that's what we notice here from our, our derivation. So first in this problem, let's take it where N one equals five. OK. And at N one equals five in the equation derived from our second equation, then T is going to be equal to five, multiplied by pi divided by five, which would just be equal to pi, which written to two significant figures would be 3.1 seconds. Likewise, for a next multiple of 510 or the next multiple of 510, that means T would be equal to 10 pi divided by five and 10 divided by five is two. So at time T would be two pi or again to two significant figures 6.3 seconds. So this tells us then that the next two consecutive times after the pendulums begin swimming swinging at which the Bobs pass, the equilibrium position simultaneously are 3.1 and 6.3 seconds. Therefore, B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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