A pendulum on a 75-cm-long string has a maximum speed of 0.25 m/s. What is the pendulum's maximum angle in degrees?

Video transcript

Welcome back, everyone. We are making observations about a Bob that is suspended from a long string that we are told is 1.2 m long. Now, the string oscillates like a pendulum where we are told that the speed of the pendulum at the equilibrium. Therefore, the maximum velocity is 1.5 m per second. And we are tasked with finding what is the maximum displacement angle of the Bob in degrees. So what are we gonna do here? Well, since it's moving like a pendulum, we can apply the formulas of simple harmonic motion to this problem here, we know that position in simple harmonic motion is given by the amplitude times the cosine of our Omo omega T plus our phase constant. And we also know that Omega is equal to the square root of gravitational acceleration divided by the length of our string here. Going back to our position equation. What I'm gonna do is I'm gonna take the derivative of both sides of this equation with respect to T in order to get velocity. What we get is that V max is equal to the negative amplitude of sine of Omega T plus our face constant using the chain rule derivative of the inside. So therefore, times omega, what this gives us is our amplitude can simply just be uh found by multiplying our angle displacement times the length of our wire. And everything else is the same Omega sine omega T plus our phase constant. However, we know that Omega is equal to the square root of G over L. So let's plug that into our equation here, plugging it into the inside of the trig function. Uh Because we also know that Omega is equal to two pi times F, that'll make every single value of T or this entire inside parentheses equal to zero. This, therefore, this term goes away. So we have that V max is equal to the negative of theta max times the length of our string, times the square root of G over L. Keep writing the wrong L there. Here you go. And what I'm gonna do is I'm actually gonna move the L on the outside under the radical sign and we can do that by simply squaring it. So we'll have negative theta max, I uh times the square root of L squared G over L which just simplifies to negative theta max. Let's see here, negative theta max times the square root of G L and going back here, right? When we plug in on or take our derivative, this negative sign is going to disappear here. Great. So then dividing both sides by square root of G over L square root of G over L. We have that our theta max is equal to V max over the square root of G over L. So let's go ahead and plug that in. We have that this is equal to 1.5 divided by the square roots of 9.8 times 1.2, which gives us 0.437 radiant. But we need this in degrees. So we're going to multiply this by pi radiant or 1 80 degrees divided by pi radiant, which gives us that our theta max is equal to 25 degrees which corresponds to our answer. Choice of c Thank you all so much for watching. I hope this video helped. We'll see you all in the next one.