A simple pendulum consists of a mass m swinging at the end...

A simple pendulum consists of a mass $m$ swinging at the end of a rod or wire of negligible mass. The figure shows a simple pendulum of length $L$. The angular displacement $\theta$ at time $t$ is given by a trigonometric expression: $$\theta(t)=A \sin (\omega t+\phi)$$ where $A, \omega, \phi$ are constants. (FIGURE CAN'T COPY) (a) Show that the function $\theta$ satisfies the equation $$\frac{d^{2} \theta}{d t^{2}}+\omega^{2} \theta=0$$ (Except for notation, this is the equation of Exercise $71 .$) (b) Show that $\theta$ can be written in the form $$\theta(t)=A \sin \omega t+B \cos \omega t$$ where $A, B, \omega$ are constants.

A simple pendulum consists of a mass $m$ swinging at the end of a rod or wire of negligible mass. The figure shows a simple pendulum of length $L$. The angular displacement $\theta$ at time $t$ is given by a trigonometric expression: $$\theta(t)=A \sin (\omega t+\phi)$$ where $A, \omega, \phi$ are constants. (FIGURE CAN'T COPY)
(a) Show that the function $\theta$ satisfies the equation $$\frac{d^{2} \theta}{d t^{2}}+\omega^{2} \theta=0$$
(Except for notation, this is the equation of Exercise $71 .$)
(b) Show that $\theta$ can be written in the form $$\theta(t)=A \sin \omega t+B \cos \omega t$$ where $A, B, \omega$ are constants.

Key Concepts

Simple Pendulum

A simple pendulum is a physical system consisting of a mass (often called a bob) attached to a light, inextensible rod or string, which swings under the influence of gravity. It is a classic example used to demonstrate principles of periodic motion and energy conservation in mechanics.

Simple Harmonic Motion

Simple harmonic motion (SHM) is a type of oscillatory motion where the restoring force is directly proportional to the displacement from equilibrium and acts in the opposite direction. Its behavior is represented by sinusoidal functions, making it an important model for studying oscillations in various physical systems, including the pendulum under small-angle approximations.

Second Order Homogeneous Differential Equations

The motion of the pendulum, when described for small angular displacements, can be modeled by a second order homogeneous differential equation with constant coefficients. This form of differential equation typically represents systems exhibiting oscillatory behavior, and its general solution involves sinusoids or exponential functions.

Trigonometric Representations of Oscillatory Motion

Oscillatory motions, such as those described by simple harmonic motion, are often expressed using trigonometric functions like sine and cosine. The displacement of the pendulum can be written in amplitude-phase form as A sin(?t + ?) or equivalently in the sum-of-sinusoids form as A sin(?t) + B cos(?t). This flexibility in representation is key for solving and analyzing problems involving periodic systems.

Transcript

00:01 Okay, here we're given the function theta t, which can be written as a times sine of omega t plus.

00:12 Okay, well, a, omega, and phi are all some constant.

00:20 And for the first part of our question, we're required to check.

00:27 Maybe we need to use this notation.

00:30 The second derivative of theta with respect to t plus omega squared times theta is equal to zero.

00:39 Okay, to check this equation, we know we need to compute the derivative of theta.

00:47 Because to get the second order derivative, we should get the first order derivative first.

00:53 And by the chair rule, we know it will be equal to a times cosine omega t plus phi times by the chair rule, d omega t plus phi dt.

01:13 Okay, so it will be equal to a times omega squared times cosine omega t plus phi.

01:24 Okay, now the second derivative will be the derivative of this term.

01:29 So the second derivative by definition will be equal to, now a times cosine squared is just a constant.

01:36 So it'll be a times omega squared.

01:40 And the derivative of cosine will be minus sine.

01:43 So that's minus sine omega t plus phi times, by the chair rule, we'll have another term, which is d omega t plus phi over dt.

02:05 So we'll get, here we don't have omega squared.

02:12 We need one omega because the derivative of this term is actually equal to omega.

02:24 Here, we q one omega can be written as this way...

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