the-figure-shows-a-pendulum-with-length-l-that-makes-a-maximum-angle-theta_0-with-the-vertical-using

Question

The figure shows a pendulum with length $ L $ that makes a maximum angle $ 	heta_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} 	heta_0 ight) $ and $ g $ is the acceleration due to gravity. If $ L = 1 m $ and $ 	heta_0 = 42^\circ $, use Simpson's Rule with $ n = 10 $ to find the period.

Foster Wisusik · Accepted Answer

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&#x27;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&#x27;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&#x27;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&#x27;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&#x27;re sorry. This should be t t should be 2.65 seconds

Problem

The intensity of light with wavelength $\lambda$ …

Problem 42 Hard Difficulty

Answer

$$
T=2.6518
$$

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Video Transcript

Calculus: Early Transcendentals

Grace H.

Anna Marie V.

Kayleah T.

Kristen K.

Integration Review - Intro

Definite vs Indefinite

$$T=2.6518$$

Calculus: Early Transcendentals

Grace H.

Anna Marie V.

Kayleah T.

Kristen K.

Integration Review - Intro

Definite vs Indefinite

Grace H.

Anna Marie V.

Kayleah T.

Kristen K.

$$
T=2.6518
$$