In Section 14–5, the oscillation of a simple pendulum (Fig. 14...

Question

In Section 14–5, the oscillation of a simple pendulum (Fig. 14–48) is viewed as linear motion along the arc length 𝓍 and analyzed via F = ma. Alternatively, the pendulum’s movement can be regarded as rotational motion about its point of support and analyzed using T = Iα . Carry out this alternative analysis and show that θ (t) = θₘₐₓ cos ( √g/ℓ t + θ) ,
where θ (t) is the angular displacement of the pendulum from the vertical at time t, as long as its maximum value is less than about .
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where θ (t) is the angular displacement of the pendulum from the vertical at time t, as long as its maximum value is less than about .

Accepted Answer

Welcome back. Everyone. In this problem. A museum exhibit features a large decorative pendulum swinging back and forth. The pendulum consists of a heavy bulb of mass M at the end of a long light rod of length. L taking into account the gravitational acceleration. G derived the equation for the angular displacement of the pendulum. Given that the maximum angular displacement is small, assume the pendulum undergoes simple harmonic motion for small displacements and use the relationship that says the torque is equal to the moment of inertia multiplied by the angular acceleration for our answer choices. A says its data marks multiplied by the cosine of GT divided by L PB says its data marks multiplied by the cosine of the square root of L divided by G multiplied by TP C says its the mas multiplied by the sine of GT divided by L and D says its the marks multiplied by the cosine of the square root of G divided by L multiplied by T plus phi no. To figure out. Or rather, our problem tells us that we want to find the equation for the angular displacement of the pendulum. And we want to make use of what we know about the torque. Now, given that our torque, some of the torque is equal to the moment of inertia multiplied by the angular acceleration. We have to ask ourselves, what does any of this have to do with the angular displacement theta? In other words, what does the torque have to do with the, what does the, the angular accelerate? Sorry, the angular acceleration have to do with the, how does it all relate? Well, for starters, OK. We know that the moment of inertia for our long light rod of or for our pendulum right is equal to ML squared. OK. Next, we also know that if we choose the AX, when we choose the axis going to the point of support of the pendulum perpendicular to the plane of the screen, the only force causing a torque on the pendulum bob is gravity. And now for small angular displacements, the restoring torque due to gravity, OK, is going to be equal to negative MG multiplied by the L multiplied by L sine theta. Now taking torque causing a counterclockwise rotation to be negative. And for small angles, we can approximate theta to be equal to sine theta, which then means we can approximate the torque. OK. Let me write it on the same line. We can approximate the torque to be equal to negative MGL theta. OK? Because sign theta is approximately equal to theta. Finally, we also know that our angular acceleration is really the second derivative of the angular displacement. In other words, theta is equal to the second derivative of the angular displacement with respect to time. So using all of these relationships in what we know about the talk in our formula for the talk. OK. Then that means that the sum of the talk negative MGL theta is going to be equal to the moment of inertia ML squared multiplied by our angular acceleration, which is the second derivative of angular displacement with respect to time. Now notice here we can cancel M from both sides. OK. We can cancel L from the left hand side and L into L squared on the right hand side leaves L so no, OK. We're left with G theta or negative G theta equal to LD square multiplied by the second derivative of the angular displacement with respect to time. And no, that tells us then that the second derivative OK. Well, you know what I'm moving a bit quickly. First, let me make sure we all see this that we can divide both sides by L OK. And divide both sides by L do not have negative G theta divided by L being equal to second derivative of the angular displacement, which now means we can rewrite this. OK. As the second derivative with respect to time plus G divided by L multiplied by theta being equal to zero. No, there's something pretty important here. OK? Because we can also recall, we make some space, we can also recall that our angular velocity is equal to the square root of G divided by L. So now that means then that G divided by L must be equal to omega squared. And with that idea, we can rewrite this as our second derivative of the angular displacement with respect to time plus omega squared theta being equal to zero. Now look closely at this equation. Does it resemble anything you've seen before? Doesn't it look like the second order differential for simple harmonic motion? Yes, it is the only difference is that no, it's for the angular displacement instead of displacement and the angular velocity instead of velocity. So now in this case, we know that for a second order derivative, the general solution, it tells us that the general solution for the angular displacement with theta max is the maximum angular displacement and Phi is the phase constant. OK. Is going to be equal to theta max multiplied by the cosine of Omega T plus Phi. Earlier we said Omega equals the square root of G divided by L. So we can replace it in our formula to be equal to theta max multiplied by the cosine of the square root of G divided by LT plus Phi. Thus, this is going to be the equation for the angular displacement of the pendulum as a function of time. Therefore, that means D is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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