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By using the expressions given below, find the ratio T_{exact}/T_{ideal} for \theta = 15^{\circ}...

Question:

By using the expressions given below, find the ratio {eq}T_{exact}/T_{ideal} {/eq} for {eq}\theta = 15^{\circ} {/eq} and for {eq}\theta = 30^{\circ} {/eq}. You should get the values close to (but not exactly equal to) 1.000, indicating that {eq}T_{ideal} {/eq} is valid at the percent level for somewhat sizable angles.

{eq}T_{ideal}= 2\Pi \sqrt{L/g} {/eq}

{eq}T_{exact}= 2\Pi \sqrt{L/g} {/eq} {{eq}{1+ (1/4)\sin ^{2}(\theta /2)} {/eq}+ ...}

Period of a Simple Pendulum:

For an idealized simple pendulum, the period of oscillation is given by {eq}T = 2\Pi\sqrt{L/g}{/eq}. However, this formula only applies for small angular displacements. For any other angular displacement ({eq}\theta{/eq}) we need to use the form of the period expressed as an infinite series, or, {eq}T = 2\Pi \sqrt{L/g}\left(1 + \frac{1^2}{2^2} \ sin^2 \ \frac{\theta}{2} + \frac{1^2 \ \cdot \ 3^2}{2^2 \ \cdot \ 4^2} \ sin^4 \ \frac{\theta}{2} +... \right){/eq}. The terms can be extended for a more precise value.

Answer and Explanation:

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Part 1:

Find the ratio {eq}\dfrac{T_{exact}}{T_{ideal}}{/eq} when {eq}\theta = 15^{\circ}{/eq}:

{eq}\begin{align} \dfrac{T_{exact}}{T_{ideal}} & =...

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Learn more about this topic:

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Pendulums in Physics: Definition & Equations

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Chapter 11 / Lesson 6
30K

After watching this lesson, you will be able to explain what a pendulum is, why it is an example of simple harmonic motion, and use equations to solve pendulum problems. A short quiz will follow.


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