Mathematically, the relationship between the restoring force of a pendulum is expressed by FR = Fg (sin ?) FR = Fg (cos ?) FR = Fg (sin ?) FR = Fg (cos ?)
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The restoring force of a pendulum is the force that brings the pendulum back to its equilibrium position. This force is always directed towards the equilibrium position and is proportional to the displacement from the equilibrium position. Show more…
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Determine the work done by this force, F, to move the pendulum from ̴̵̶̷̸̡̢̧̨̛̖̗̘̙̜̝̞̟̠̣̤̥̦̩̪̫̬̭̮̯̰̱̲̳̹̺̻̼͇͈͉͍͎̈́̂̃̄̅̆̇̈̉̊̋̌̍̎̏̐̑̒̓̔̽̾̿̀́͂̓̈́͆͊͋͌̕̚ͅ͏͓͔͕͖͙͚͐͑͒͗͛ͣͤͥͦͧͨͩͪͫͬͭͮͯ͘͜͟͢͝͞͠͡ͰͱͲͳʹ͵Ͷͷͺͻͼͽ;Ϳ΄΅Ά·ΈΉΊΌΎΏΐΑΒΓΔΕΖΗΘΙΚΛΜΝΞΟΠΡΣΤΥΦΧΨΩΪΫάέήίΰαβγδεζηθικλμνξοπρςστυφχψωϊϋόύώϏϐϑϒϓϔϕϖϗϘϙϚϛϜϝϞϟϠϡϢϣϤϥϦϧϨϩϪϫϬϭϮϯϰϱϲϳϴϵ϶ϷϸϹϺϻϼϽϾϿθ = 0 to θ = θ₀. Express your answer in terms of the variables m, l, θ₀, and appropriate constants. Enter the argument of the trigonometric function in parenthesis.
Keerti J.
$\bullet\bullet\bullet$ The forces acting on a simple pendulum are shown in $\nabla$ Fig. $13.26 .$ (a) Show that, for the small angle approximation $(\sin \theta \approx \theta),$ the force producing the motion has the same form as Hooke's law. (b) Show by analogy with a mass on a spring that the period of a simple pendulum is given by $T=2 \pi \sqrt{L / g}$. [Hint: Think of the effective spring constant.]
Period of a pendulum A standard pendulum of length $L$ that swings under the influence of gravity alone (no resistance) has a period of $$ T=\frac{4}{\omega} \int_{0}^{\pi / 2} \frac{d \varphi}{\sqrt{1-k^{2} \sin ^{2} \varphi}} $$ where $\omega^{2}=g / L, k^{2}=\sin ^{2}\left(\theta_{0} / 2\right), g=9.8 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration due to gravity, and $\theta_{0}$ is the initial angle from which the pendulum is released (in radians). Use numerical integration to approximate the period of a pendulum with $L=1 \mathrm{m}$ that is released from an angle of $\theta_{0}=\pi / 4$ rad.
Integration Techniques
Numerical Integration
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