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The figure shows a pendulum with length $L$ that makes a maximum angle $\theta_0$ with the vertical. Using Newton's Second Law, it can be shown that the period $T$ (the time for one complete swing) is given by$$T = 4 \sqrt{\frac{L}{g}} \int_0^{\frac{\pi}{2}} \frac{dx}{\sqrt{1 - k^2 \sin^2 x}}$$where $k = \sin \left (\frac{1}{2} \theta_0 \right)$ and $g$ is the acceleration due to gravity. If $L = 1 m$ and $\theta_0 = 42^\circ$, use Simpson's Rule with $n = 10$ to find the period.

$$T=2.6518$$

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Anna Marie V.

Campbell University

Heather Z.

Oregon State University

Kristen K.

University of Michigan - Ann Arbor

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Video Transcript

Okay, so this question wants us to approximate the period of a pendulum using this expression So T is equal to four times a squared about all over g times this elliptic integral. So to evaluate this, it wants us to use Simpson's rule with an equal to 10. But first we consult for some constants to make things simpler later. So it says that l is one meter and we know that G is 9.81 times are integral from zero to pi over too of DX over square root of one minus well, k equals sign of 42 over two degrees, which equals 0.358 four. So weaken sub that in for our value of K to get are integral. So t equals four times the square root of one over 9.1 times the integral from zero pi over too, goes Well, let's just call this nasty expression f of X. So only have to do is approximate this integral now. So to do this, it wants us to take an equal to attend. So Delta X is well, we end at pi over to when we started zero and we want 10 sub intervals. So Delta X equals pi over 20. So s a 10 equals pi over 20 divided by three. So pie over 60 times f of zero plus four times af of our first value pirate 20 plus two times f of two pi over 20 plus f of 10 pie over 20. What you'll notice is pi over too. And then we also know that l equals four times squared of one over 9.81 times s a 10 earliest. That's our approximation. So if we plug in these values for our function because remember we defined f of X earlier Here is this expression. So now we can just plug this in using our calculator and see that l is approximately equal to two points. 65 We're sorry. This should be t t should be 2.65 seconds

University of Michigan - Ann Arbor

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Integration Techniques

Anna Marie V.

Campbell University

Heather Z.

Oregon State University

Kristen K.

University of Michigan - Ann Arbor

Lectures

Join Bootcamp