The Pendulum Is Reaching Its Limit Before It Begins to Swing Back Again.

Pendulums

Simple harmonic motion

P.U.P.A Gilbert , in Physics in the Arts (3rd Edition), 2022

11.4 Pendulum

A pendulum is a mass fastened to a cord. The top of the cord is a stock-still signal, the lesser mass is free to swing back and forth, as shown in Fig. 11.1 . Examples of pendulums are a kid on a swing, or a hammock betwixt two trees, or a cherry on its stalk. When a forcefulness moves it from its equilibrium position, the pendulum starts oscillating with uncomplicated harmonic move. The restoring force that pulls the pendulum dorsum to its equilibrium position is gravity, or, more accurately, the component of gravity along the deportation x direction, which for modest pendulum angles is:

Effigy xi.1. The position x of a pendulum at different times t. (a) $\mathrm{At}t=0\text{s},x=+1.5\text{m}$ , You pulled the pendulum to the right and released it. If you count as positive the displacement to the correct of the equilibrium position, also called a rest position, the initial displacement is $x=+\mathrm{1.v}\text{thou}$ . (b) $t=\mathrm{i}\text{s},x=+1.0\text{thou}$ , Gravity is the forcefulness that pulls the mass downward, but the string keeps it from falling directly down. Instead, the mass is pulled back toward the equilibrium position at the bottom of the pendulum swing. The force of gravity accelerates the movement of the mass. (c) $t=\mathrm{two}\text{due south},\mathrm{ten}=0.0\text{yard}$ , The pendulum is at the equilibrium position, just it keeps moving considering of inertia. In that location is no breaking force to finish information technology. (d) $t=3\text{s},x=-\mathrm{one.0}\text{m}$ , Now the strength of gravity pulls the mass dorsum toward the equilibrium position, information technology therefore slows down, or decelerates, the motion of the mass, just the pendulum keeps going until (east) $t=\mathrm{iv}\text{s},x=-\mathrm{ane.v}\text{1000}$ , the continuing downward pull of gravity eventually stops the mass at $x=-\mathrm{1.five}\text{thou}$ . But gravity continues to pull the mass toward the equilibrium position, thus the oscillation keeps going back and forth periodically.

$F=-\frac{mg}{L}x$

Where m is the mass hanging from the pendulum string, chiliad  =   nine.eight   m/s² is the acceleration of gravity, and 50 is the length of the string. The quantity $\frac{\mathrm{yard}m}{\mathrm{Fifty}}$ is a feature of each pendulum, and tin be abbreviated equally the pendulum constant $k=\frac{\mathrm{1000}g}{L}$ .

Therefore, the restoring strength exerted by gravity on a pendulum is:

$F=-k\mathrm{ten}\text{where}k=\frac{mg}{L}$

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Angular Motility

Paul Davidovits , in Physics in Biology and Medicine (Fifth Edition), 2019

4.5 Concrete Pendulum

The uncomplicated pendulum shown in Fig. 4.three is non an adequate representation of the swinging leg because information technology assumes that the total mass is located at the end of the pendulum while the pendulum arm itself is weightless. A more realistic model is the physical pendulum, which takes into account the distribution of weight along the swinging object (see Fig. 4.five). It tin exist shown (run into [24] ^one ) that under the force of gravity the period of oscillation $T$ for a physical pendulum is

Effigy 4.5. The physical pendulum.

(4.13) $T=2\pi \sqrt{\frac{I}{\mathit{Wr}}}$

Here $I$ is the moment of inertia of the pendulum around the pivot indicate $O$ (run across Appendix A); $\mathrm{West}$ is the total weight of the pendulum, and $r$ is the distance of the center of gravity from the pivot point. (The expression for the flow in Eq. 4.13 is once more strictly correct only for small angular deportation.)

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OSCILLATORY MOTION

George B. Arfken , ... Joseph Priest , in University Physics, 1984

Example three Measurement of chiliad with a Simple Pendulum

A simple pendulum (Figure 14.10) tin can be used to determine the acceleration of gravity, m, at a particular location. The expression for the period is

Figure xiv.x. The period T of a simple pendulum of length 50 may be used to measure g, the acceleration of gravity.

$T=2\text{π}\sqrt{\frac{l}{g}}$

Thus we have

$\mathrm{1000}=\frac{4{\text{π}}^{\mathrm{two}}l}{T^{\mathrm{two}}}$

for the acceleration of gravity in terms of the measured catamenia, T, and length, l, of the pendulum.

A pendulum that is 1.00 chiliad in length is observed to crave 100.60 s to execute 50 complete oscillations. These information give a period

$\begin{array}{ll}T\hfill & =\frac{100.60}{50}\hfill \\ \hfill & =2.01\text{s}\hfill \end{array}$

Thus nosotros summate

$\begin{array}{ll}g\hfill & =\frac{4{\text{π}}^{\mathrm{two}}\times 1}{{\left(2.01\right)}^{\mathrm{two}}}\hfill \\ \hfill & =9.77\text{m}/{\text{s}}^2\hfill \end{array}$

for the acceleration of gravity.

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LIQUID SLOSHING

R. Ibrahim , in Encyclopedia of Vibration, 2001

The pendulum model

For the pendulum model shown in Figure 3 , each pendulum represents 1 sloshing mode. A rigidly attached mass is called to represent the effect of the liquid that moves in unison with the tank as a frozen mass. By comparing the expression of hydrodynamic force due to lateral excitation of the tank and the total force expression due to the system of pendulums and the rigid mass, the following model parameters are obtained:

Figure three. Pendulum equivalent modeling.

(16) $\begin{array}{l}{\mathrm{thou}}_n=\frac{\mathrm{viii}\rho b{a}^2}{{(2n-\mathrm{i})}^3{\pi }^3}tanh(\mathrm{ii}\mathrm{north}-1)\frac{\pi h}{a}\\ m_0=\rho abh-\sum _{n=1}^{\infty }{m}_{\mathrm{due\; north}}\\ {\mathcal{l}}_{\mathrm{north}}=\frac{a}{(2n-1)\pi }coth(\mathrm{two}\mathrm{due\; north}-\mathrm{one})\frac{\pi h}{a}\\ H_n={\mathcal{l}}_n+\frac{h}{2}-\frac{2a}{(\mathrm{ii}n-1)\pi }tanh(\mathrm{two}\mathrm{due\; north}-\mathrm{one})\frac{\pi h}{a}\end{array}$

Similarly, the resulting moment near the heart of mass yields the following constants. The mass moment of inertia of the frozen mass is:

(17a) $I_0=I_F-m_0{H}_0^2-\stackrel{\mathrm{\infty }}{\sum _{n=1}}{m}_{\mathrm{northward}}{\left(H_{\mathrm{northward}}-{\mathrm{\ell }}_{\mathrm{due\; north}}\right)}^{\mathrm{two}}$

I _F is the fluid mass moment of inertia given by the expression:

(17b) $\begin{array}{l}{I}_F=\{\rho abh\frac{h^{\mathrm{ii}}}{12}+\frac{a^{\mathrm{ii}}}{16}-2a^{\mathrm{two}}\sum _{\mathrm{due\; north}=\mathrm{i}}^{\infty }\frac{16}{{(2\mathrm{north}-\mathrm{i})}^{\mathrm{iv}}{\pi }^4}\\ \times \left[\mathrm{ane}-\frac{2a}{(\mathrm{ii}n-\mathrm{ane})\pi h}tanh(\mathrm{two}\mathrm{northward}-1)\frac{\pi h}{2a}\right]\end{array}$

With reference to Figure iii, m _north is the nth pendulum mass, and fifty _northward represents the nth pendulum length. H ₀ is the altitude of the tank center of mass to the center of mass of the frozen fluid portion of mass m _n . H _n is the distance of the nth pendulum support point to the tank center of mass.

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Oscillations and Waves

Ruslan P. Ozerov , Anatoli A. Vorobyev , in Physics for Chemists, 2007

Example E2.v

A physical pendulum consists of a rod and a hoop of masses k _rod = 3thousand ₁ and m _hoop = thousand ₁; the length of the rod is l = 1 m. The horizontal axis of oscillation Oz is perpendicular to the rod and passes it at its center O. Determine the oscillation menstruation of such a pendulum. The rest of the definitions are given in Figure E2.5.

Solution: The period of oscillation is expressed by eq. (2.4.14), T = 2π $\sqrt{\frac{I}{mg{l}_c}}$ .

To find the period, we must starting time cull a reference frame (axis x), mark a zero position on it (run into Figure E2.5) and find the MI of parts of the pendulum, I _z,1 is the MI of the rod and I _z,2 is the MI of the hoop and the full MI is I. It is besides necessary to find the altitude l _c between the oscillation axis (bespeak O) and the CM. The MI of the physical pendulum relative to the oscillation axis is the sum of I ₁ (the MI of a rod) and I ₂ (the MI of the hoop both relative to the same centrality I = I ₁ + I ₂).

The rod MI is I _ane = chiliad ₁ 50 ²/4 (considering the rod mass is threek _ane and the axis passes through the centre of the rod). The MI of the hoop is the sum of the MI of the hoop itself (first item) and the addition from the parallel centrality theorem (2nd particular):

$I_2={\mathrm{thousand}}_1{\left(\frac{\ell }{4}\right)}^2+m_1{\left(\frac{3\ell }{\mathrm{iv}}\right)}^2=\frac{\mathrm{five}{m}_1{\ell }^2}{8}.$

The total MI of the pendulum is the sum I = $\frac{m_1{l}^2}{4}+\frac{5m_1{l}^{\mathrm{two}}}{8}=\frac{\mathrm{seven}{m}_1{l}^2}{8}$

The altitude

${\ell }_c=\frac{\Sigma m_i{x}_i}{\Sigma {\mathrm{1000}}_i}=\frac{\mathrm{three}{g}_1\times 0+m_1\left(\mathrm{iii}\ell /\mathrm{four}\right)}{3m_{\mathrm{one}}+m_{\mathrm{one}}}=\frac{\frac{3}{4}{m}_{\mathrm{i}}\ell }{4{\mathrm{chiliad}}_1}=\frac{3}{16}\ell .$

(In order to simplify calculation of l _c it is useful to mark nothing on axis x at the aforementioned level as bespeak O; in this example the CM coordinate is simultaneously l _c). Obtaining these preliminary results we can place all the values under the square root:

$T=2\pi \sqrt{\frac{\frac{7}{8}{\mathrm{chiliad}}_1{\ell }_{\text{c}}^2}{4{\mathrm{thousand}}_{1}g\frac{3}{16}{\ell }_{\text{c}}}}=\mathrm{ii}\pi \sqrt{\frac{7{\ell }_{\text{c}}}{\mathrm{six}g}},$

therefore, T = 2.17 sec.

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OSCILLATORY Motion

George B. Arfken , ... Joseph Priest , in International Edition University Physics, 1984

xiv.3 Applications of Simple Harmonic Motility

The period of unproblematic harmonic motility depends on the concrete properties of the oscillating arrangement and its surroundings. A precise measurement of the period is possible past counting many vibrations, and often provides a dynamic method of determining characteristics of the system or of its environs. For instance, using Eq. 14.v

${\omega }^{\mathrm{ii}}=\frac{k}{m}$

and Eq. xiv.14

$\omega =\frac{2\pi }{T}$

nosotros can write the period for a horizontal spring-mass system as

(xiv.21) $T=2\pi {\left(\frac{m}{k}\right)}^{\mathrm{½}}$

If we use a calibrated spring (Section iii.one), and so the value of k is known. Thus a measurement of T can be used to measure thousand. This experiment is of particular interest because it is simply the inertial mass that is involved. This technique was used in Skylab for measuring mass. Common mass measurements using balances or vertical springs involve the forcefulness of gravity.

In Example 2 the measured period of a seconds pendulum (two s) together with the known value of g enabled u.s.a. to decide the pendulum length fifty. We can reverse this experiment and use a pendulum of measured length with a measured period to infer a value for g Using Eq. fourteen.6

${\omega }^2=\frac{g}{l}$

and Eq. xiv.14, we can express g in terms of l and T:

(14.22) $g=\frac{4{\pi }^{2}l}{T^2}$

In Newton's twenty-four hour period the pendulum was used to obtain values for g at unlike latitudes over the surface of the earth. Newton's conjecture that the globe was an oblate spheroid was confirmed in this way

For a simple pendulum, T tin can be determined with accuracy, but it is not possible to obtain an equally accurate value of fifty. This in plow limits the accuracy of the values of thou obtained by using a uncomplicated pendulum. The state of the art in the measurement of g is represented by the method used at the National Bureau of Standards. An object is projected upward in a vacuum and immune to retrace its path as suggested in Figure 14.9. It rises, passing levels I and II at the times $t_1$ and $t_{\mathrm{two}},$ and so falls, passing levels Ii and I at times $t_{\mathrm{iii}}$ and $t_{\mathrm{four}}.$ Precise measurements of $h$ , the vertical distance between levels I and II, and of the four times are made, permitting the ciphering of an accurate value for four (encounter Chapter four, Problem 44).

Figure 14.9. A particle traveling straight up passes level I at t ₁ and level II at t _2. On the way back down the particle retraces its path, passing level 2 at t₃ and level I at t _4. The distance h between levels and the iv times t₁ , t₂ , t_iii , t ₄ tin can exist measured accurately. These data are used to determine an accurate value for 1000, the dispatch of gravity.

Example 3

Measurement of one thousand with a Simple Pendulum

A simple pendulum (Figure 14.x) can be used to determine the acceleration of gravity, k, at a detail location. The expression for the menstruation is

$T=2\pi \sqrt{\frac{l}{k}}$

Figure xiv.ten. The period T of a unproblematic pendulum of length $\mathrm{fifty}$ may be used to measure thou, the acceleration of gravity. and 2 at the times t₁ and t_two , and then falls, passing levels Two and I at times t ₃ and t _4. Precise measurements of h, the vertical altitude between levels I and II, and of the four times are made, permitting the ciphering of an accurate value for m (encounter Chapter 4, Problem 44).

Thus we have

$\mathrm{grand}=\frac{\mathrm{four}{\pi }^2\mathrm{50}}{T^2}$

for the acceleration of gravity in terms of the measured menstruation, T, and length, l, of the pendulum.

A pendulum that is 1.00 m in length is observed to require 100. 60 s to execute 50 complete oscillations. These data give a period

$\begin{array}{l}T=\frac{100.60}{50}\\ =\mathrm{two.01}\text{s}\end{array}$

Thus we calculate

$\begin{array}{l}\mathrm{yard}=\frac{4{\pi }^2\times 1}{{\left(2.01\right)}^{\mathrm{ii}}}\\ =9.77{\text{thou/due south}}^2\end{array}$

for the acceleration of gravity.

Measurement of the Moment of Inertia

A torsion pendulum (see Sections six.6 and 26.3) consists of a body suspended on the end of a wire or rod that can exist prepare into rotational oscillations. The period of such oscillations depends on the moment of inertia of the system. A torsion pendulum can exist used to measure the unknown moment of inertia of an object. A vertical torsion pendulum, arranged to mensurate the moment of inertia of a standing person about an axis through the center of mass, is shown in Figure fourteen.11. We will testify how a measurement of T for torsional vibrations tin be used to obtain I for the person.

Figure fourteen.11. A torsion pendulum is an object gratuitous to rotate when suspended by a vertical wire or rod. If the wire is twisted and released, it unwinds and the object undergoes rotational oscillations. Hither the torsion pendulum period is used to measure the moment of inertia of a person.

First, we must calibrate the torsion pendulum. A known torque, τ, is applied to twist the rod until information technology comes to equilibrium at the bending θ. Nosotros presume that Hooke'south law holds for the rod, then that the ratio of applied torque to twist angle is abiding. This assumption permits us to write

(14.23) $\tau =-\beta \theta$

The minus sign is needed because τ is a restoring torque. A static measurement of τ and θ establishes the value of the twist constant of the rod, β.

Side by side the pendulum—without the person in place—is twisted through the angle θ and released. If $I_{\circ }$ is the moment of inertia of the empty pendulum, the rotational course of Newton's second law enables us to write

$\tau =-\beta \theta =I_o\alpha$

or

$\alpha =-\left(\frac{\beta }{I_o}\right)\theta$

showing that the pendulum executes SHM. Thus nosotros have for the flow

(fourteen.24) $\begin{array}{lll}{T}_{\circ }=2\pi {\left(\frac{I_{\circ }}{\beta }\right)}^{1/2}& \text{or}& I_{\circ }\end{array}=\frac{\beta T_{\circ }^{\mathrm{two}}}{4{\pi }^{\mathrm{two}}}$

Finally, the person climbs onto the pendulum platform, and the pendulum is once more twisted through the bending θ and released. If I_p is the moment of inertia of the person lone, we can follow the same procedure to write

$\tau =-\beta \theta =\left(I_o+I_p\right)\alpha$

or

$\alpha =-\left(\frac{\beta }{I_o+I_p}\right)\theta$

showing that we once more accept SHM. Thus we can write for the flow

(xiv.25) $\begin{array}{lll}T=2\pi {\left(\frac{I_o+I_p}{\beta }\right)}^{\mathrm{½}}& \text{or}& I_o\end{array}+I_p=\frac{\beta T^{\mathrm{ii}}}{\mathrm{iv}{\pi }^{\mathrm{ii}}}$

Substituting the value of I₀ obtained in Eq. xiv.24 into Eq. 14.25, we can solve for the moment of inertia of the person, I_p ,

(14.26) $I_p=\frac{\beta }{\mathrm{four}{\pi }^{\mathrm{ii}}}\left(T^2-T_{\mathrm{o}}^2\right)$

in terms of the three measured quantities β, T, and T_o. By having the person presume various positions we can measure out any desired moment of inertia of the person.

Example 4

Moment of Inertia of a Person

Allow us use the torsion pendulum method to determine the moment of inertia of a person. The person stands vertically with the symmetry line of the apparatus passing through his or her middle of mass. We measure β = 125 N · m/rad, T _o = two.000 southward, and T = 2.152 s. What is the person's moment of inertia?

Using Eq. 14.26 nosotros have

$\begin{array}{l}{I}_p=\frac{125}{4{\pi }^2}\left[{\left(2.152\right)}^2-{2.000}^2\right]\\ =2.00\text{kg}\cdot {\text{m}}^{\text{2}}\end{array}$

for the moment of inertia of the person. Note that, because of the counterfoil in Eq. 14.26, T _o and T must exist measured to 4 significant figures in lodge to obtain I_p to three significant figures.

Questions

half-dozen.

If oscillations through large angles are permitted, the period of a elementary pendulum depends on amplitude. Will the period corresponding to large amplitude be greater than or less than the period corresponding to small aamplitude? Give a physical argument in back up of your reply.

vii.

Explain why it is of import for the person to stand at the exact eye of the platform in Case 4.

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EQUILIBRIUM OF RIGID BODIES

George B. Arfken , ... Joseph Priest , in University Physics, 1984

Questions

five.

A pendulum bob of weight W is shown in Figure 3.22 hanging on the end of a cord at an angle. Give an argument based on the first condition of equilibrium showing that the pendulum brawl tin can't be in equilibrium.

Figure iii.22. A simple pendulum.

6.

2 forces act on a person continuing on level ground—his weight W, acting downward; and the normal force P, acting upwards. In club for the person to be in translational equilibrium, the first condition of equilibrium requires that P = Westward. Explain why P and W do not establish an activeness-reaction pair of forces.

vii.

Is the answer given in Case 5 reasonable? Use some of the solution checks discussed in this section to make up one's mind.

8.

Explain why a particle experiencing merely one force cannot be in equilibrium.

9.

A particle is in equilibrium nether the influence of two forces. Explain why the forces must exist equal in magnitude and oppositely directed.

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Pendulum, resonance and molecular highly excited vibration

GUOZHEN WU , in Nonlinearity and Anarchy in Molecular Vibrations, 2005

ix.i Pendulum

The motion of a pendulum is a very basic and of import physical miracle. When the amplitude of a pendulum is very pocket-size, its frequency is a constant and independent of the amplitude. It is a simple harmonic move. When its amplitude is larger, its frequency is no longer a constant and depends on the amplitude or energy. As the vibrational frequency is related to its energy, we telephone call the system a nonlinear ane.

The potential energy of a pendulum V is proportional to ane− cosθ = 2 sinⁱⁱ θ/2. Hither, θ is the respective angle. The functional form of V(θ) is shown in Fig.9.1. As θ is smaller, Five(θ)~θⁱⁱ. It is a parabola. Every bit θ is close to ± π, V(θ) will non vary much. This feature of V(θ) leads to the quantized levels as shown in Fig.9.1. Equally the energy is smaller, the energy spacings are almost the aforementioned (having the character of a simple harmonic movement). When θ is close to ± π, the energy spacing becomes smaller.

Fig.9.ane. The potential of a pendulum. The horizontal lines in the potential valley show the quantized levels. Due to the nonlinear event, the energy spacing is less for higher levels.

The phase space of the move of a pendulum is shown in Fig.9.2. Phase space is the relation of action versus angle. (The relation of J versus θ. J is the action, θ is the angle.) As the energy of a pendulum is not enough for θ to exceed π (or − π), the motion is stable and periodic. In this situation, the phase construction is a closed ellipse. Its heart is a stable fixed point, corresponding to the stationary land. When the energy increases, θ will exceed the range of (-π, π), corresponding to the rotation. The bespeak b at the intersection of the separatrices separating these ii types of motion is an unstable fixed betoken. It is a hyperbolic point. Around indicate b, at that place are both stable and unstable spaces. We annotation that as movement is along the separatrix, reaching to point b and stopping there, the time required is infinite. This is because when motion is closer to indicate b, the speed becomes less and approaches zero!

Fig.9.2. The phase diagram of the movement of a pendulum. J is the action and θ is the angle. Curves correspond to the various quantized levels. Arrows denote the directions of motion. A, A′ evidence the opposite rotations. Southward is the separatrix. a (a′) and b points are stable and unstable fixed points, respectively.

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Applications of Higher Order Differential Equations

Martha L. Abell , James P. Braselton , in Introductory Differential Equations (5th Edition), 2018

5.5 The Pendulum Problem

Suppose that a mass yard is fastened to the stop of a rod of length L, the weight of which is negligible. (Run across Fig. 5.nineteen.) We want to determine an equation that describes the motility of the mass in terms of the displacement $\theta (t)$ , which is measured counterclockwise in radians from the vertical centrality shown in Fig. five.xix. This is possible if we are given an initial position and an initial velocity of the mass. A force diagram for this situation is shown in Fig. five.twentyA.

Figure 5.19. A swinging pendulum.

Effigy 5.20. Two force diagrams for the swinging pendulum.

Notice that the forces are determined with trigonometry using the diagram in Fig. 5.20A. In this instance, $\cos \theta =\mathrm{yard}k/\mathrm{10}$ and $\sin \theta =m\mathrm{one\; thousand}/y$ , so we obtain the forces

$x=gm\cos \theta \text{and}y=mg\sin \theta ,$

which are indicated in Fig. five.20B.

The momentum of the mass is given by in $\mathrm{thousand}d\mathrm{southward}/dt$ , so the rate of change of the momentum is

$\frac{d}{dt}\left(m\frac{ds}{dt}\right)=m\frac{d^{2}s}{d{t}^2},$

where southward represents the length of the arc formed by the motion of the mass. So, because the forcefulness $m\mathrm{1000}\sin \theta$ acts in the contrary management of the motion of the mass, we have the equation

$m\frac{d^{2}s}{d{t}^2}=-mm\sin \theta .$

(Observe that the force $\mathrm{1000}g\cos \theta$ is commencement by the force of constraint in the rod, then mg and $m\mathrm{thou}\cos \theta$ cancel each other in the sum of the forces.) Using the human relationship from geometry between the length of the arc, the length of the rod, and the angle θ, $s=L\theta$ , nosotros have the relationship

$\frac{d^{\mathrm{two}}s}{d{t}^2}=\frac{d^{\mathrm{two}}}{d{t}^2}(L\theta )=L\frac{d^2\theta }{d{t}^{\mathrm{ii}}}.$

The deportation $\theta (t)$ satisfies

$\begin{array}{cc}\hfill & mL\frac{d^{\mathrm{ii}}\theta }{d{t}^2}=-mg\sin \theta \text{or}\hfill \\ \hfill & mL\frac{d^2\theta }{d{t}^2}+\mathrm{thousand}k\sin \theta =0,\hfill \end{array}$

which is a nonlinear equation. However, because we are merely concerned with small displacements, we note from the Maclaurin serial for $\sin \theta$ ,

$\sin \theta =\theta -\frac{1}{3!}{\theta }^3+\frac{1}{5!}{\theta }^5-\frac{1}{\mathrm{vii}!}{\theta }^7+\cdots ,$

that for pocket-sized values of θ, $\sin \theta \approx \theta$ . Therefore, we obtain the linear equation $mL{d}^{\mathrm{two}}\theta /d{t}^{\mathrm{two}}+m\mathrm{1000}\theta =0$ or $d^{\mathrm{ii}}\theta /d{t}^{\mathrm{two}}+(g/\mathrm{50})\theta =0$ , which approximates the original problem. If the initial deportation (position of the mass) is given past $\theta (0)={\theta }_0$ and the initial velocity (the velocity with which the mass is gear up into motion) is given by ${\theta }^{\prime }(0)=v_0$ , we have the initial value problem

(5.9) $\frac{d^2\theta }{d{t}^2}+\frac{g}{L}\theta =0,\theta (0)={\theta }_0,\frac{d\theta }{dt}(0)=5_0$

to find the deportation function $\theta (t)$ .

Suppose that ${\omega }^2=g/L$ so that the differential equation becomes $d^2\theta /d{t}^{\mathrm{ii}}+{\omega }^2\theta =0$ . Therefore, functions of the form

$\theta (t)=c_{\mathrm{i}}\cos \omega t+c_2\sin \omega t,$

where $\omega =\sqrt{g/L}$ , satisfy the equation $d^2\theta /d{t}^2+g/\mathrm{50}\theta =0$ . When we use the conditions $\theta (0)={\theta }_0$ and ${\theta }^{\prime }(0)=v_0$ , we find that the role

(v.10) $\theta (t)={\theta }_0\cos \omega t+\frac{v_0}{\omega }\sin \omega t$

satisfies the equation likewise as the initial deportation and velocity conditions. As we did with the position function of spring-mass systems, we can write this function as a cosine function that includes a phase shift with

(5.11) $\begin{array}{cc}\hfill & \theta (t)=\sqrt{{\theta }_0^{\mathrm{ii}}+\frac{v_0^2}{{\omega }^{\mathrm{ii}}}}\cos (\omega t-\varphi ),\text{where}\hfill \\ \hfill & \varphi ={\cos }^{-1}\left(\frac{{\theta }_0}{\sqrt{{\theta }_0^{\mathrm{ii}}+{\mathrm{five}}_0^2/{\omega }^2}}\right)\hfill \end{array}$

and $\omega =\sqrt{g/L}$ .

Note that the period of $\theta (t)$ is $T=\mathrm{two}\pi /\omega =\mathrm{ii}\pi \sqrt{L/g}$ .

Case 5.17

Determine the deportation of a pendulum of length $\mathrm{Fifty}=8$ feet if $\theta (0)=0$ and ${\theta }^{\prime }(0)=2$ . What is the period? If the pendulum is office of a clock that ticks once for each time the pendulum makes a complete swing, how many ticks does the clock brand in one minute?

Solution: Considering $g/L=32/8=4$ , the initial value problem that models this situation is

$\frac{d^{\mathrm{two}}\theta }{d{t}^2}+4\theta =0,\theta (0)=0,\frac{d\theta }{dt}(0)=\mathrm{ii}.$

A general solution of the differential equation is $\theta (t)=c_1\cos 2t+c_2\sin 2t$ , so application of the initial conditions yields the solution $\theta (t)=\sin \mathrm{two}t$ . The period of this function is

$T=\mathrm{ii}\pi \sqrt{\frac{L}{g}}=2\pi \sqrt{\frac{8\text{ ft}}{32\text{ ft}/{\text{southward}}^{\mathrm{two}}}}=\pi \text{ s}.$

(Notice that we can use our knowledge of trigonometry to compare the period with $T=2\pi /\mathrm{ii}=\pi$ .) Therefore, the number of ticks made by the clock per minute is calculated with the conversion

$\frac{1\text{ rev}}{\pi \text{ s}}\times \frac{1\text{ tick}}{\mathrm{ane}\text{ rev}}\times \frac{60\text{ south}}{1\text{ min}}\approx \mathrm{xix.ane}\text{ ticks}/\text{min}.$

Hence the clock makes approximately nineteen ticks in one infinitesimal.  □

How is motion afflicted if the length of the pendulum in Example 5.17 is changed to $\mathrm{50}=\mathrm{iv}$ ?

If the pendulum undergoes a damping strength that is proportional to the instantaneous velocity, the forcefulness due to damping is given by $F_R=bd\theta /dt$ . Incorporating this forcefulness into the sum of the forces acting on the pendulum, we obtain the nonlinear equation $L{d}^2\theta /d{t}^2+bd\theta /dt+g\sin \theta =0$ . Again, using the approximation $\sin \theta \approx \theta$ for small values of θ, we employ the linear equation $L{d}^2\theta /d{t}^{\mathrm{two}}+bd\theta /dt+g\theta =0$ to approximate the state of affairs. Therefore, nosotros solve the initial value problem

(5.12) $\begin{array}{cc}\hfill & L\frac{d^2\theta }{d{t}^2}+b\frac{d\theta }{dt}+\mathrm{yard}\theta =0,\theta (0)={\theta }_0,\hfill \\ \hfill & \frac{d\theta }{dt}(0)=v_0\hfill \end{array}$

to find the deportation role $\theta (t)$ .

Example 5.18

A pendulum of length $\mathrm{Fifty}=8/5\text{ ft}$ is subjected to the resistive strength $F_R=32/\mathrm{five}d\theta /dt$ due to damping. Determine the displacement function if $\theta (0)=\mathrm{ane}$ and ${\theta }^{\prime }(0)=\mathrm{two}$ .

Solution: The initial value problem that models this situation is

$\begin{array}{cc}\hfill & \frac{\mathrm{viii}}{\mathrm{five}}\frac{d^2\theta }{d{t}^{\mathrm{ii}}}+\frac{32}{\mathrm{v}}\frac{d\theta }{dt}+32\theta =0,\theta (0)=1,\hfill \\ \hfill & \frac{d\theta }{dt}(0)=2.\hfill \end{array}$

Simplifying the differential equation, we obtain $d^2\theta /d{t}^{\mathrm{ii}}+4d\theta /dt+20\theta =0$ , which has characteristic equation $r^2+4r+20=0$ with roots $r_{1,\mathrm{two}}=-2\pm 4i$ . A full general solution is $\theta (t)=e^{-\mathrm{ii}t}(c_1\cos 4t+c_2\sin 4t)$ . Awarding of the initial conditions yields the solution $\theta (t)=e^{-\mathrm{two}t}(\cos 4t+\sin \mathrm{four}t)$ . Nosotros graph this solution in Fig. 5.21. Observe that the damping causes the displacement of the pendulum to decrease over time. □

Figure 5.21. Plot of $\theta (t)=e^{-\mathrm{two}t}(\cos 4t+\sin \mathrm{four}t)$ .

When does the object in Example 5.xviii first pass through its equilibrium position? What is the maximum deportation from equilibrium?

In many cases, we tin can utilise computer algebra systems to obtain authentic approximations of nonlinear problems.

Instance 5.19

Use a figurer algebra organization to approximate the solutions of the nonlinear problems (a) $\frac{d^2\theta }{d{t}^2}+\sin \theta =0$ , $\theta (0)=0$ , $\frac{d\theta }{dt}(0)=\frac{1}{2}$ and (b) $\frac{\mathrm{eight}}{5}\frac{d^2\theta }{d{t}^2}+\frac{32}{5}\frac{d\theta }{dt}+32\theta =0$ , $\theta (0)=1$ , $\frac{d\theta }{dt}(0)=\mathrm{two}$ . Compare the results to the corresponding linear approximations.

Solution: (a) We bear witness the results obtained with a typical computer algebra arrangement in Fig. 5.22. We come across that equally t increases, the gauge solution, $\theta =\frac{\mathrm{one}}{2}\sin t$ , becomes less accurate. However, for small values of t, the results are nearly identical.

Figure five.22. (A) The numerical solution to the nonlinear problem is in night red and the exact solution to the linear approximation is in light red. (B) The accented value of the difference between the two approximations.

(b) The exact solution to the respective linear approximation is obtained in Example five.18. We testify the results obtained with a typical figurer algebra system in Fig. five.23. In this case, nosotros see that the fault diminishes as t increases. (Why?) □

Figure 5.23. (A) The numerical solution to the nonlinear problem is in dark red and the verbal solution to the linear approximation is in light ruby. (B) The absolute value of the deviation betwixt the two approximations.

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Free energy gap and tunneling

Charles P. PooleJr., ... Richard J. Creswick , in Superconductivity (Third Edition), 2014

F Analogues of Josephson junctions

Josephson tunneling involves a quantum miracle that is difficult to grasp intuitively. This is particularly true when we try to picture how the total electric current flowing through a Josephson junction depends on the stage deviation of the electron pairs on either side of the junction. The differential equation for this stage difference ϕ happens to exist the same as the differential equation for the rotational movement of a driven pendulum. We will draw this motion and so relate it to the Josephson junction.

Consider a uncomplicated pendulum consisting of a mass 1000 fastened to a pivot past a massless rod of length R. If a constant torque τ is practical past a motor, it will motility the mass through an angle ϕ, as shown in Fig. xiii.44. Nosotros know from our report of mechanics that the force of gravity interim on the mass grand produces a restoring torque mgR sinϕ. For a relatively small practical torque the pendulum assumes an equilibrium position at the bending given by

Figure 13.44. Pendulum model of a Josephson junction showing the counterclockwise restoring torque mgR sinϕ arising from the presence of a clockwise applied torque τ.

(13.66) $\tau =\mathrm{grand}gR\sin \varphi \left(\frac{\mathrm{d}\varphi }{\mathrm{d}t}=0\right),$

as indicated in Fig. 13.45b. The greater the torque, the larger the angle ϕ. There is a critical torque τ _c indicated in Fig. 13.45c for the angle ϕ=π/2,

Figure 13.45. Pendulum (a) with no applied torque, τ=0, (b) with the torque τ=½ mgR, and (c) with the critical torque applied, τ _c=mgR.

(13.67) ${\tau }_{\mathrm{c}}=mgR.$

If the practical torque exceeds this disquisitional value, the pendulum will continue its motion beyond the angle ϕ=π/2 and rotate continuously as long as the practical torque τ>τ _c operates. The move is fast at the bottom and boring at the top, corresponding to a large angular velocity ω=dϕ/dt at the bottom and a pocket-sized ω at the peak. For a big torque, τ≫mgR, the average angular velocity of the motion ⟨ω⟩ increases linearly with the torque, reaching a limit determined by retarding drag forces coming from, for example, the viscosity η of the air or mechanical friction. The elevate force is causeless to be proportional to the angular velocity ω, and is written every bit ηω.

The dependence of the boilerplate angular velocity on the applied torque is shown in Fig. 13.46. We meet from the effigy that ω remains zero equally the torque τ is increased until the critical value τ _c=mgR of Eq. (xiii.67) is reached. Beyond this indicate, ω jumps to a finite value and continues to rising in the way we have already described. If the torque is now decreased downward from a large magnitude, in one case it passes the critical value of Eq. (xiii.67), the pendulum will have sufficient kinetic energy to keep it rotating for torques below τ _c, as indicated in the figure. The torque must be reduced much further, down to the value ${\tau \prime }_{\mathrm{c}},$ earlier friction begins to dominate and motion stops, as indicated in the effigy. Thus we have hysteresis of motion for low applied torques, and no hysteresis for high torques. If nosotros compare Fig. xiii.46 with Fig. thirteen.36b, nosotros see that the torque–angular velocity feature curve of the driven pendulum has the same shape equally the current–voltage characteristic of the Josephson junction.

Figure 13.46. Relationship between the average athwart velocity of the pendulum ⟨ω⟩ and the practical torque τ. For low applied torques the pendulum oscillates and the average velocity is zip, whereas at high torques, τ>τ _c, motion is continuous with ⟨ω⟩ proportional to τ. Notation the hysteresis for increasing and decreasing torques.

The correspondence between the driven pendulum and a Josephson junction can be demonstrated by writing down a differential equation that governs the motility of the pendulum, setting the practical torque τ equal to the charge per unit of alter of the athwart momentum Fifty,

(13.68) $\frac{\mathrm{d}}{\mathrm{d}t}L=\mathrm{chiliad}{R}^2\frac{\mathrm{d}\omega }{\mathrm{d}t},$

and then adding the restoring and damping torques,

(thirteen.69) $\tau =\mathrm{grand}{R}^2\frac{{\mathrm{d}}^2\varphi }{\mathrm{d}{t}^{\mathrm{ii}}}+\eta \frac{\mathrm{d}\varphi }{\mathrm{d}t}+\mathrm{one\; thousand}kR\sin \varphi ,$

where mR ⁱⁱ is the moment of inertia; hither we have made use of the expression ω=dϕ/dt. This equation is mathematically equivalent to its Josephson counterpart (13.51), so we tin can make the following identifications:

Applied current	I	↔	τ	Applied torque
Average voltage term	Five(2e/ħ)=dϕ/dt	↔	ω=dϕ/dt	Boilerplate angular velocity
Stage difference	ϕ	↔	ϕ	Athwart deportation
Capacitance term	ħC/2e	↔	mR two	Moment of inertia
Conductance term	ħG/iidue east	↔	η	Viscosity
Critical current	I c	↔	mgR	Critical torque

This analogue has been establish useful in the study of the behavior of Josephson junctions.

Some other mechanical device that illustrates Josephson junction-type behavior is the washboard counterpart sketched in Fig. 13.47, in which a particle of mass yard moves down a sloped sinusoidal path in a viscous fluid, passing through regularly spaced minima and maxima along the way.

Figure 13.47. Washboard counterpart of the Josephson junction showing a particle of mass yard descending forth a sloped wavy path in a pasty fluid.

Electrical analogues have been proposed (Bak and Pedersen, 1973; Hamilton, 1972; Hu and Tinkham, 1989; cf. Goodrich and Srivastava, 1992; Goodrich et al., 1991) that do not give as much insight into Josephson junction behavior every bit the mechanical analogues, notwithstanding, they are useful for studying the beliefs of Josephson junctions when the parameters are varied.

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