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Chapter 4: Problem 3

A simple pendulum consists of a particle of mass \(m\) hung on a massless string of length \(l\) in the earth's gravitational field. Its period for small- amplitude motion is then \(T=2 \pi(l / g)^{1 / 2}\). Find by a perturbation calculation how the period is aitered if the pendulum actually has a small moment of inertia \(I_{0}\) about its center of mass, which we continue to take at a distance \(l\) from the point of support. The moment of inertia \(I_{0}\) includes contributions from the mass of the string and from the finite dimensions of the particle.

Short Answer

Expert verified
Answer: The period of the pendulum is altered by the introduction of the small moment of inertia \(I_0\). The new formula for the altered period is \(T = 2π(\frac{l}{g(1+ε)})^{\frac{1}{2}}\), taking into account the contributions from the mass of the string and the finite dimensions of the particle.

Step by step solution

01

1. Equation of Motion

To find the equation of motion for the pendulum, we use the angular displacement \(\theta\) around the point of support as a coordinate. The torque \(τ\) about the point of support is given by \(τ = -mglsin(θ)\). For small values of \(θ\), we can approximate the sine function as \(sin(θ)≈θ\). So, the torque becomes \(τ ≈ -mglθ\). Now, we can use the equation of motion for rotation: \(τ=Iα\), with \(I\) being the moment of inertia of the pendulum and \(α\) being the angular acceleration. Here, \(I = ml^2 + I_0\). Therefore, the equation of motion becomes \((ml^2 + I_0)α = -mglθ\).
02

2. Apply Perturbation Calculation

We need to apply the perturbation calculation to the equation of motion derived in step 1, considering the contribution of \(I_0\). To do so, let \(I_0 = ε(ml^2)\), where \(ε\) is a small coefficient. So, the equation of motion becomes \((ml^2 + ε(ml^2))α = -mglθ\). Divide both sides by \(ml^2\) to obtain \((1 + ε)α = -\frac{g}{l}θ\)
03

3. Finding the Altered Period

Using the new equation of motion \((1 + ε)α=-\frac{g}{l}θ\), we can rewrite it as a differential equation: \((1 + ε)\frac{d^2θ}{dt^2}=-\frac{g}{l}θ\). The solution of this equation is of the form \(θ(t)=Acos(\omega t) + Bsin(\omega t)\), where \(A\) and \(B\) are constants, and \(ω\) is the angular frequency of the oscillation. Plugging the solution back into our equation, we obtain \((1 + ε)(-\omega^2) = -\frac{g}{l}\). Solving for \(ω\), we get \(ω = \sqrt{\frac{g}{l(1+ε)}}\). Now, the period of the oscillation is related to the angular frequency through \(T = \frac{2π}{ω}\). So, substituting ω back in, we get the altered period as \(T = 2π(\frac{l}{g(1+ε)})^{\frac{1}{2}}\). Thus, the period of the pendulum is altered by the introduction of the small moment of inertia \(I_0\). The new formula for the altered period is \(T = 2π(\frac{l}{g(1+ε)})^{\frac{1}{2}}\), taking into account the contributions from the mass of the string and the finite dimensions of the particle.

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Key Concepts

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

Perturbation Calculation
Perturbation calculation is a mathematical technique used to approximate solutions to problems that are too complex to solve directly. It's particularly useful when a system experiences a slight modification that impacts its behavior. In the context of a simple pendulum with a slight moment of inertia, the perturbation is represented by the introduction of a small coefficient, typically denoted as \(\theta\).

By assuming the perturbation parameter \(\epsilon\) is much less than 1, we treat the moment of inertia as a minor alteration to the system and analyze its effects. Perturbation calculations allow us to adjust the classical period formula to account for the small moment of inertia without solving the entire complex system from scratch. This method can simplify the problem and give us an estimate of how the period of oscillation is 'altered' due to the additional factor of the moment of inertia.
Moment of Inertia
The moment of inertia, symbolized as \(I\), is a measure of an object's resistance to changes in its rotation. Think of it as the rotational equivalent of mass for linear motion. For a pendulum, the moment of inertia is crucial as it influences how the pendulum swings and its period.

In our case, the moment of inertia is affected by not only the mass of the pendulum bob but also the mass distribution of the string and the bob's finite dimensions. When introduced into calculations for the pendulum's period, a small moment of inertia \(I_0\) adds to the pendulum's total moment of inertia, slightly altering the system's dynamics and, consequently, its period.
Equation of Motion
An equation of motion describes how a physical system evolves over time under the influence of forces. For rotational motion around a fixed axis, as with our simple pendulum, the equation of motion connects the torque exerted on the system and the resultant angular acceleration.

Using the simple relation \(\tau = I\alpha\), where \(\tau\) is the torque, \(I\) is the moment of inertia, and \(\alpha\) is the angular acceleration, we derive an equation that describes the pendulum's swing. The simplicity of this approach makes it easier for students to relate a physical concept (torque), a motion property (angular acceleration), and inertia, thereby deepening their understanding of rotational dynamics.
Angular Displacement
Angular displacement refers to the angle through which a point or line has been rotated in a specified direction around a specific axis. For a swinging pendulum, angular displacement measures how far the pendulum swings from its equilibrium position and is given by the angle \(\theta\).

Angular displacement is key to understanding the motion of oscillating systems like pendulums. Small displacements can be linearized using the small-angle approximation, which simplifies calculations by making the assumption that \(\sin(\theta) \approx \theta\) when \(\theta\) is expressed in radians and is very small. This linearization is part of the backdrop to our solution, making the resulting math more approachable.
Small-Angle Approximation
The small-angle approximation is a simplification used when dealing with oscillatory systems like the simple pendulum. It relies on the fact that for small angles (measured in radians), the sine of the angle is approximately equal to the angle itself: \(\sin(\theta) \approx \theta\).

This approximation is extremely useful because it can transform trigonometric functions, which can complicate equations, into simple polynomial ones. In the textbook problem, by making this approximation, we can linearize the equation, which significantly eases the calculation of the new period altered by the introduction of the moment of inertia. This approximation underpins many areas in physics and engineering where precision can be balanced with simplification for small angles.

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Most popular questions from this chapter

An exponential horn whose area is given by (4.3.3) makes an excellent impedance transformer. Let us investigate the properties of such a horn whose length is an integral number of half-wavelengths, so that \(k_{1} l=n \pi\). In the notation of Sec. \(4.2\), the wave on the horn is $$ \xi=e^{-\alpha x}\left(\breve{A}_{4} e^{-j \boldsymbol{\kappa}_{1} \boldsymbol{z}}+\breve{A}_{-} e^{j \boldsymbol{\mu}_{1} \boldsymbol{x}}\right) e^{j \omega t} . $$ Compute the characteristic impedance \(\breve{Z}(x)=F / \xi\) at any position \(x\) on the horn. Set it equal to the load impedance \(\bar{Z}_{t}\) at \(x=l=\pi / n k_{1}\) and show that the impedance at \(x=0\) is then \(\breve{Z}(0)=\left(S_{0} / S_{l}\right) \breve{Z}_{l}\), where \(S_{0}\) is the area of the horn at \(x=0\) and \(S_{l}\) is the area at \(x=l\).

Show that longitudinal waves on a rod of circular section whose radius \(a\) is small compared with a wavelength obey the wave equation $$ Y \frac{\partial^{2} \xi}{\partial x^{2}}=\rho_{0}\left(\frac{\partial^{2} \xi}{\partial l^{2}}-\frac{1}{2} \sigma^{2} a^{2} \frac{\partial^{4} \xi}{\partial x^{2} \partial l^{2}}\right) $$ when radial motion is taken into account through Poisson's ratio \(\sigma .\) Generalize the equation to a rod of noncircular cross section. Show that (4.4.24) is consistent with (4.4.23).

Consider a medium in which \(c_{\text {phase }}=A \omega^{n}\), where \(A\) and \(n\) are constants. Show that \(c_{\text {eroup }}=c_{\text {pasae }} /(1-n)\). For what values of \(n\) is the dispersion normal? Anomalous?

Two long waveguides having different characteristic impedances \(Z_{1}\) and \(Z_{2}\) are joined by a quarter-wave section having the characteristic impedance \(\left(Z_{1} Z_{2}\right)^{1 / 2}\). A wave \(\xi_{1}=\) \(A_{1} e^{i\left(\omega t-x_{1} x\right)}\) is traveling toward the coupling section. Assume that there is a reflected wave \(\xi_{1}=\check{B}_{1} e^{i\left(\omega t+x_{1} x\right)}\), that there are waves going in both directions in the coupling section, and that there is a wave \(\xi_{2}=X_{2} e^{i\left(\omega t-x_{1} x\right)}\) transmitted into the other waveguide. By satisfying boundary conditions at the two junctions, show that \(\breve{B}_{1}=0\) and that \(\breve{A}_{2}=A_{1}\left(Z_{1} / Z_{2}\right)^{1 / 2}\), which is the value one may get very easily from (4.2.10) by power considerations alone.

Show that (4.9.18) for the second-order frequency calculation can be put in the form $$ \left(\frac{\omega_{21}}{\omega_{01}}\right)^{2}=1-G_{2}-\frac{1}{4} \sum_{n=2}^{\infty} n\left(\frac{n+1}{n-1} G_{n+1}^{2}-\frac{n-1}{n+1} G_{n-1}^{2}\right) $$ where $$ G_{n}=\frac{2}{l} \int_{0}^{l} \ln S(x) \cos \frac{n \pi x}{l} d x . $$ This form is more suitable for practical calculations.
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