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A simple pendulum consists of a point mass m suspended by a (weightless) cord or rod of length l, as shown, and swinging in a vertical plane under the action of gravity. Show that for small oscillations (small θ ), both θ and x are sinusoidal functions of time, that is, the motion is simple harmonic. Hint: Write the differential equation F=ma for the particle m. Use the approximation sinθ=θ for small θ, and show that θ=Asinωt is a solution of your equation. What are A and ω?

Short Answer

Expert verified
θ(t)=Asin(glt) and x(t)=lAsin(glt). The motion is simple harmonic with ω=gl.

Step by step solution

01

Set up the differential equation using Newton's second law

Consider the forces acting on the mass. The restoring force due to gravity acting along the arc is F=mgsinθ. According to Newton's second law, this force is equal to the mass times the acceleration along the arc mx¨. Since x=lθ, we have mx¨=mlθ¨.
02

Use the small-angle approximation

For small oscillations, sinθθ. Therefore, substituting this approximation into the force equation gives mlθ¨=mgθ.
03

Simplify the differential equation

Divide both sides of the equation by ml to isolate θ¨: θ¨+glθ=0
04

Identify the simple harmonic motion equation

The resulting differential equation for θ, θ¨+glθ=0 is a standard form of the simple harmonic motion equation θ¨+ω2θ=0, where ω=gl.
05

Solve for θ(t)

The general solution to this differential equation is θ(t)=Asin(ωt), where A is the amplitude.
06

Define the angular frequency ω

From the previous steps, the angular frequency ω is ω=gl. Thus, θ(t)=Asin(glt) represents the sinusoidal function of time for θ.
07

Relate the arc length x to θ

Since x=lθ, substituting θ(t) gives x(t)=lAsin(glt), showing that x is also a sinusoidal function of time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton's Second Law is a fundamental principle in physics that states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. For the simple pendulum, the force acting on the point mass is the restoring force due to gravity. This force can be expressed as F=mgsin(θ), where mgsin(θ) represents the component of the gravitational force pulling the mass back towards its equilibrium point. By Newton's Second Law, this force is also equal to the mass of the pendulum bob times its acceleration along the arc, which gives us the equation: mx¨=mgsin(θ). Since x=lθ, we can write the equation as: m(lθ)¨=mgsin(θ). Simplifying this expression leads us to the differential equation that describes the motion of the pendulum.
Small-Angle Approximation
The Small-Angle Approximation is a key concept when analyzing the motion of a simple pendulum. It assumes that for small angles (measured in radians), the sine of the angle is approximately equal to the angle itself. This simplifies the mathematical analysis of the pendulum's motion. In other words, for small oscillations of the pendulum, we can use: sin(θ)θ. This approximation is valid when θ is in the range of a few degrees. By applying the small-angle approximation to our differential equation, we can replace sin(θ) with θ. This results in a simplified differential equation: mlθ¨=mgθ. Dividing both sides by ml, we get: θ¨+glθ=0. This is a key step towards understanding the simple harmonic motion of the pendulum.
Differential Equation
A Differential Equation is an equation that involves the derivatives of a function. For the simple pendulum, we form a differential equation to describe its motion. From Newton's Second Law and applying the small-angle approximation, we obtained the equation: mlθ¨=mgθ. After simplification, this becomes: θ¨+glθ=0. This is a second-order linear differential equation, which is characteristic of simple harmonic motion (SHM). The general form of SHM differential equation is: x¨+ω2x=0, where ω is the angular frequency. By comparing our pendulum's differential equation with the SHM equation, we can identify ω and confirm that the motion of the pendulum is indeed simple harmonic for small angles.
Angular Frequency
Angular Frequency (ω) is a measure of how quickly something oscillates in simple harmonic motion. For the simple pendulum, we derived the differential equation: θ¨+glθ=0. Comparing this with the standard form of the simple harmonic motion equation: x¨+ω2x=0, we can see that ω2=gl. Therefore, the angular frequency ω of the pendulum is: ω=gl. This shows how the length l of the pendulum and the acceleration due to gravity g influence the frequency of its oscillations. The period of the pendulum, which is the time it takes to complete one full oscillation, is related to the angular frequency by: T=2πω=2πlg. This gives us a complete understanding of how the pendulum behaves over time.

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Most popular questions from this chapter

Find the average value of the function on the given interval. Use equation (4.8) if it applies. If an average value is zero, you may be able to decide this from a quick sketch which shows you that the areas above and below the x axis are the same. 1ex on (0,1)

Show that if f(x) has period p, the average value of f is the same over any interval of length p. Hint: Write aa+pf(x)dx as the sum of two integrals (a to p, and p to a+p) and make the change of variable x=t+p in the second integral.

(a) Find the exponential Fourier transform of f(x)=e|x| and write the inverse transform. You should find 0cosαxα2+1dα=π2e|x| (b) Obtain the result in (a) by using the Fourier cosine transform equations (12.15) (c) Find the Fourier cosine transform of f(x)=1/(1+x2). Hint: Write your result in (b) with x and α interchanged.

Let f(x) and g(α) be a pair of Fourier transforms. Show that df/dx and iαg(α) are a pair of Fourier transforms. Hint: Differentiate the first integral in (12.2) under the integral sign with respect to x. Use (12.23) to show that α|g(α)|2dα=12πif¯(x)ddxf(x)dx. Comment: This result is of interest in quantum mechanics where it would read, in the notation of Problem 12.35:p|ϕ(p)|2dp=ψ(x)(ih2πddx)ψ(x)dx.

(a) Prove that 0π/2sin2xdx=0π/2cos2xdx by making the change of variable x=12πt in one of the integrals. (b) Use the same method to prove that the averages of sin2(nπx/l) and cos2(nπx/l) are the same over a period.

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