What is the restoring force on a displaced simple pendulum?

Gravity's tangential component, directed toward equilibrium: $F=-mg\sin\theta$ when displacement away from equilibrium is positive.

How does the small-angle restoring force relate to arc displacement?

For small angles, $F\approx -(mg/L)s$, where $s$ is the arc displacement.

What is the period of a simple pendulum for small oscillations?

$T=2\pi\sqrt{L/g}$, where $L$ is the length to the bob's center of mass and $g$ is local gravitational acceleration.

Which properties determine a simple pendulum's period for small swings?

Its length and local gravitational acceleration; the period is independent of bob mass and nearly independent of amplitude.

How can pendulum measurements be used to estimate local gravity?

Measure the pendulum length $L$ and period $T$, then calculate $g=4\pi^2L/T^2$.

General physicsLesson 2 of 2 in the course Oscillatory Motion and Waves

Simple Pendulum: Restoring Force, Period, and Measuring Gravity

5 flashcards · 5 quiz questions · 5 match pairs · 5 fill-in sentences · English

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Summary

A simple pendulum is a useful model for understanding oscillatory motion. In this lesson, we examine the forces on a pendulum bob and see why its motion is approximately simple harmonic when its swing is small.

A simple pendulum consists of a small, dense object called the bob, suspended from a fixed point by a light string or wire of length $L$ . When the bob is displaced from its equilibrium position and released, it swings along an arc.

All flashcards

What is the restoring force on a displaced simple pendulum?: Gravity's tangential component, directed toward equilibrium: $F=-mg\sin\theta$ when displacement away from equilibrium is positive.
How does the small-angle restoring force relate to arc displacement?: For small angles, $F\approx -(mg/L)s$ , where $s$ is the arc displacement.
What is the period of a simple pendulum for small oscillations?: $T=2\pi\sqrt{L/g}$ , where $L$ is the length to the bob's center of mass and $g$ is local gravitational acceleration.
Which properties determine a simple pendulum's period for small swings?: Its length and local gravitational acceleration; the period is independent of bob mass and nearly independent of amplitude.
How can pendulum measurements be used to estimate local gravity?: Measure the pendulum length $L$ and period $T$ , then calculate $g=4\pi^2L/T^2$ .

Quiz questions

For small oscillations, which equation gives the period $T$ of a simple pendulum with length $L$ in local gravitational field $g$ ?Answer: $T=2\pi\sqrt{L/g}$
For small swings, what happens to a simple pendulum's period if the bob's mass is increased while its length and location remain unchanged?Answer: The period remains unchanged.
For small oscillations at the same location, what happens to the period if a pendulum's length is increased by a factor of four?Answer: The period doubles.
Which equation estimates local gravitational acceleration from a simple pendulum's length $L$ and measured period $T$ ?Answer: $g=4\pi^2L/T^2$
Why is timing several complete pendulum oscillations and dividing by their number useful?Answer: It reduces the effect of reaction-time error on the period estimate.

Based on College Physics 2e by Paul Peter Urone, Roger Hinrichs (OpenStax), licensed CC BY 4.0. Adapted into a summary, flashcards and quizzes.

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