imtoken安卓版2.9.9|pendulum

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2024-03-07 20:30:02

PENDULUM中文(简体)翻译：剑桥词典

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pendulum 在英语-中文（简体）词典中的翻译

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/ˈpen.dʒəl.əm/ us

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[ C ] a device consisting of a weight on a stick or thread that moves from one side to the other, especially one that forms a part of some types of clocks

摆，摆锤;（尤指）钟摆

The pendulum in the grandfather clock swung back and forth.

落地钟的钟摆来回摆动。

The spaceship's jets were fired periodically to dampen a side-to-side pendulum motion that had developed.

宇宙飞船的喷射装置周期性地喷射，以减缓飞船出现的左右摇摆现象。

[ S ] a change, especially from one opinion to an opposite one

摇摆不定的事态（或局面）;（尤指）观点的摇摆不定

As so often in education, the pendulum has swung back to the other extreme and testing is popular again.

人们的观点又回到了另一个极端，考试又流行起来了，这种摇摆不定的现象在教育领域经常出现。

(pendulum在剑桥英语-中文（简体）词典的翻译 © Cambridge University Press)

pendulum的例句

pendulum

To some degree the periods represent "swings of the regulatory pendulum" 8 but, as will become clear, these swings are not along just one dimension.

来自 Cambridge English Corpus

Without knowing more about the particulars - for instance, about engineers'concepts of pendulums - one cannot dare to give a definitive answer in this case.

来自 Cambridge English Corpus

A useful mechanical analogy is developed by considering the effects of gravity modulation on a simple pendulum.

来自 Cambridge English Corpus

My wife tells me how much she hates pendulums and that she prefers balances.

来自 Hansard archive

该例句来自Hansard存档。包含以下议会许可信息开放议会许可v3.0

The microphones are pulled back, switched on, and released over the speaker, and gravity causes them to swing back and forth as pendulums.

来自 Wikipedia

该例句来自维基百科，在CC BY-SA许可下可重复使用。

The first affects our considerations insofar as we will calculate accumulated phase differences for pendulums of substantially different amplitudes.

来自 Cambridge English Corpus

A simple mechanical example of parametric resonance is the excitation of a simple pendulum by periodic vertical motion of its point of support.

来自 Cambridge English Corpus

A simple pendulum analogy, the stability characteristics of which are known, is shown to apply to the fluid layer system.

来自 Cambridge English Corpus

示例中的观点不代表剑桥词典编辑、剑桥大学出版社和其许可证颁发者的观点。

pendulum的翻译

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擺，擺錘, （尤指）鐘擺, 搖擺不定的事態（或局面）…

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a type of food similar to a hamburger but made without meat, by pressing together small pieces of vegetables, seeds, etc. into a flat, round shape

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pendulum - 搜索词典

ulum - 搜索词典 Rewards网页图片视频学术词典地图更多航班我的必应笔记本pendulum美 [ˈpendʒələm] 英 [ˈpendjʊləm] n.钟摆网络灵摆；摇锤；摆锤复数：pendulums 同义词n.weight,bob,plumb,swing权威英汉双解英汉英英网络释义pendulum显示所有例句n.1.钟摆a long straight part with a weight at the end that moves regularly from side to side to control the movement of a clockIn education, the pendulum has swung back to traditional teaching methods.教育界又恢复了传统教学法。the pendulum of public opinion舆论的转变n.1.(钟等的)摆2.动摇的人[物];犹豫不决的人3.吊烛架n.1.a long thin bar with a weight at the lower end that swings from side to side, usually in order to keep a clock working2.used for talking about the tendency of an opinion or situation to change from one position or condition to its opposite1.钟摆钟字的解释---在线新华字典 ... 钟馗〖 ZhongKui(ChungK’uei)〗钟摆〖 pendulum〗钟表〖 clocksandwatches〗 ... xh.5156edu.com|基于933个网页2.灵摆灵摆(pendulum)的意思是悬垂物,也就是落有重物的绳线。灵摆是一种很大众化的占卜工具,只要有线跟重物就可以进行占卜。club.qingdaonews.com|基于505个网页3.摇锤英语单词常见词根 - 豆丁网 ... compulsion 强迫, 强制 pendulum 钟摆, 摇锤 pension 养老金, 退休金 ... www.docin.com|基于111个网页4.摆锤硬度计量名词术语及定义 - 豆丁网 ... 旋转轴线( axis of rotation) 摆锤( pendulum) 锤刃( s t r i k i n g e d g e ) ... www.docin.com|基于49个网页5.摆，钟摆【考研单词速记】（私） ... behalf n. 利益 1001 ▲ pendulum n. 摆,钟摆 661 invisible a. 看不见的,无形的 6912 ... www.douban.com|基于48个网页6.振动体医学词汇（P）--上海翻译公司-021-58377363 ... pencil magnet 笔形吸铁器 pendulum 摆（锤），振动体 penetrameter X 射线 …www.2008translation.com|基于32个网页7.摆动摄影词典(300条) 数码相机知识 ZOL术语 ... 暗箱 Camera obscura 摆动 Pendulum 摆动镜头 Swing lens ... detail.zol.com.cn|基于23个网页更多释义收起释义例句释义:全部全部,钟摆钟摆,灵摆灵摆,摇锤摇锤,摆锤摆锤类别:全部全部,口语口语,书面语书面语,标题标题,技术技术来源:全部全部,字典字典,网络网络难度:全部全部,简单简单,中等中等,难难更多例句筛选收起例句筛选1.But my overriding fear was that the pendulum would swing to the other extreme if I skipped a night.但我更怕的是，假若逃过了今晚，欲望的钟摆就会摆向另一个极端。www.douban.com2.When the base of the plume moved, the rest of the feather acted as if it were a spring pendulum -- a fairly standard physics equation.当羽毛的底座移动的时候，羽毛的其他部分会按照弹簧摆的运动方式来运动——这是个颇为标准的物理方程。dictsearch.appspot.com3.The control method of Inverted Pendulum has been used in the military, aerospace, general industrial robots and industry process control.其控制方法在军工、航天、机器人领域和一般工业过程中都有着广泛的用途。www.fabiao.net4.The Pendulum Type. The waist moves slightly, which is often found in the dance of the Miao nationality.腰部的动律小，形成上身轻微的摆动，如苗族的舞蹈。dict.ebigear.com5.The cultural pendulum may be beginning to swing the other way in the direction of common sense and a more civil society.文化的钟摆可能正开始朝另一个方向摆动，一个有着共识和更文明的社会的方向。bb.wwenglish.org6.As travelers, we also seem to be on a pendulum, shifting back and forth through space, and often end up at hubs of transit.像旅行者，我们也好像在一个钟摆，来回移动穿过空间，和经常结束在运输的中心。www.mflady.com7.If I double the mass of my bob at the end of a pendulum, then the vertical component of the tension will also double.如果把钟摆末端，物体重量加倍，那么张力的垂直分力，也会加倍。open.163.com8.But in 2011, at a time of global economic uncertainty and with the U. S. embroiled in three wars, the pendulum has swung the other way.但是在2011年，在这个全球经济具有强大的不确定性的时刻，并且伴随着美国已经卷入了3场战争的形势下，事态已经发生了改变。dongxi.net9.Like magic out of a Harry Potter novel, such a metal could contract on command, or swing back and forth like a pendulum.像这种仿佛出自《哈利·波特》小说中的魔法一般的金属可以按照指令收缩，或者像钟摆那样前后摇摆。www.bing.com10.Now the pendulum is beginning to swing back as local governments attempt to minimise job losses in their own back yards.如今，由于当地政府试图将本地的失业率降至最低，钟摆开始回摆。www.ftchinese.com12345© 2024 Microsoft隐私声明和 Cookie法律声明广告帮

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15: OscillationsUniversity Physics I - Mechanics, Sound, Oscillations, and Waves (OpenStax){ }{ "15.01:_Prelude_to_Oscillations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "15.02:_Simple_Harmonic_Motion" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "15.03:_Energy_in_Simple_Harmonic_Motion" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "15.04:_Comparing_Simple_Harmonic_Motion_and_Circular_Motion" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "15.05:_Pendulums" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "15.06:_Damped_Oscillations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "15.07:_Forced_Oscillations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "15.E:_Oscillations_(Exercises)" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "15.S:_Oscillations_(Summary)" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()" }{ "00:_Front_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "01:_Units_and_Measurement" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "02:_Vectors" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "03:_Motion_Along_a_Straight_Line" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "04:_Motion_in_Two_and_Three_Dimensions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "05:_Newton\'s_Laws_of_Motion" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "06:_Applications_of_Newton\'s_Laws" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "07:_Work_and_Kinetic_Energy" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "08:_Potential_Energy_and_Conservation_of_Energy" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "09:_Linear_Momentum_and_Collisions" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "10:_Fixed-Axis_Rotation__Introduction" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "11:__Angular_Momentum" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "12:_Static_Equilibrium_and_Elasticity" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "13:_Gravitation" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "14:_Fluid_Mechanics" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "15:_Oscillations" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "16:_Waves" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "17:_Sound" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "18:_Answer_Key_to_Selected_Problems" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()", "zz:_Back_Matter" : "property get [Map MindTouch.Deki.Logic.ExtensionProcessorQueryProvider+<>c__DisplayClass230_0.b__1]()" }Mon, 12 Sep 2022 23:37:00 GMT15.5: Pendulums40654065admin{ }AnonymousAnonymous2falsefalse[ "article:topic", "Pendulums", "authorname:openstax", "simple pendulum", "physical pendulum", "torsional pendulum", "license:ccby", "showtoc:no", "program:openstax", "licenseversion:40", "source@https://openstax.org/details/books/university-physics-volume-1" ][ "article:topic", "Pendulums", "authorname:openstax", "simple pendulum", "physical pendulum", "torsional pendulum", "license:ccby", "showtoc:no", "program:openstax", "licenseversion:40", "source@https://openstax.org/details/books/university-physics-volume-1" ]https://phys.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fphys.libretexts.org%2FBookshelves%2FUniversity_Physics%2FUniversity_Physics_(OpenStax)%2FBook%253A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)%2F15%253A_Oscillations%2F15.05%253A_Pendulums

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$ \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } $ $ \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} $$\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$ $\newcommand{\id}{\mathrm{id}}$ $ \newcommand{\Span}{\mathrm{span}}$ $ \newcommand{\kernel}{\mathrm{null}\,}$ $ \newcommand{\range}{\mathrm{range}\,}$ $ \newcommand{\RealPart}{\mathrm{Re}}$ $ \newcommand{\ImaginaryPart}{\mathrm{Im}}$ $ \newcommand{\Argument}{\mathrm{Arg}}$ $ \newcommand{\norm}[1]{\| #1 \|}$ $ \newcommand{\inner}[2]{\langle #1, #2 \rangle}$ $ \newcommand{\Span}{\mathrm{span}}$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$

Learning ObjectivesThe Simple PendulumExample $\PageIndex{1}$: Measuring Acceleration due to Gravity by the Period of a PendulumSolutionExercise $\PageIndex{1}$Physical PendulumExample $\PageIndex{2}$: Reducing the Swaying of a SkyscraperSolutionTorsional PendulumExample $\PageIndex{3}$: Measuring the Torsion Constant of a StringSolution

Learning Objectives

State the forces that act on a simple pendulum

Determine the angular frequency, frequency, and period of a simple pendulum in terms of the length of the pendulum and the acceleration due to gravity

Define the period for a physical pendulum

Define the period for a torsional pendulum

Pendulums are in common usage. Grandfather clocks use a pendulum to keep time and a pendulum can be used to measure the acceleration due to gravity. For small displacements, a pendulum is a simple harmonic oscillator.

The Simple Pendulum

A simple pendulum is defined to have a point mass, also known as the pendulum bob, which is suspended from a string of length L with negligible mass (Figure $\PageIndex{1}$). Here, the only forces acting on the bob are the force of gravity (i.e., the weight of the bob) and tension from the string. The mass of the string is assumed to be negligible as compared to the mass of the bob.

Figure $\PageIndex{1}$: A simple pendulum has a small-diameter bob and a string that has a very small mass but is strong enough not to stretch appreciably. The linear displacement from equilibrium is s, the length of the arc. Also shown are the forces on the bob, which result in a net force of −mgsin$\theta$ toward the equilibrium position—that is, a restoring force.

Consider the torque on the pendulum. The force providing the restoring torque is the component of the weight of the pendulum bob that acts along the arc length. The torque is the length of the string L times the component of the net force that is perpendicular to the radius of the arc. The minus sign indicates the torque acts in the opposite direction of the angular displacement:

\[\begin{split} \tau & = -L (mg \sin \theta); \\ I \alpha & = -L (mg \sin \theta); \\ I \frac{d^{2} \theta}{dt^{2}} & = -L (mg \sin \theta); \\ mL^{2} \frac{d^{2} \theta}{dt^{2}} & = -L (mg \sin \theta); \\ \frac{d^{2} \theta}{dt^{2}} & = - \frac{g}{L} \sin \theta \ldotp \end{split}\]

The solution to this differential equation involves advanced calculus, and is beyond the scope of this text. But note that for small angles (less than 15°), sin $\theta$ and $\theta$ differ by less than 1%, so we can use the small angle approximation sin $\theta$ ≈ $\theta$. The angle $\theta$ describes the position of the pendulum. Using the small angle approximation gives an approximate solution for small angles,

\[\frac{d^{2} \theta}{dt^{2}} = - \frac{g}{L} \theta \ldotp \label{15.17}\]

Because this equation has the same form as the equation for SHM, the solution is easy to find. The angular frequency is

\[\omega = \sqrt{\frac{g}{L}} \label{15.18}\]

and the period is

\[T = 2 \pi \sqrt{\frac{L}{g}} \ldotp \label{15.19}\]

The period of a simple pendulum depends on its length and the acceleration due to gravity. The period is completely independent of other factors, such as mass and the maximum displacement. As with simple harmonic oscillators, the period T for a pendulum is nearly independent of amplitude, especially if $\theta$ is less than about 15°. Even simple pendulum clocks can be finely adjusted and remain accurate.

Note the dependence of T on g. If the length of a pendulum is precisely known, it can actually be used to measure the acceleration due to gravity, as in the following example.

Example $\PageIndex{1}$: Measuring Acceleration due to Gravity by the Period of a Pendulum

What is the acceleration due to gravity in a region where a simple pendulum having a length 75.000 cm has a period of 1.7357 s?

Strategy

We are asked to find g given the period T and the length L of a pendulum. We can solve T = 2$\pi$L g for g, assuming only that the angle of deflection is less than 15°.

Solution

Square T = 2$\pi \sqrt{\frac{L}{g}}$ and solve for g: $$g = 4 \pi^{2} \frac{L}{T^{2}} ldotp$$

Substitute known values into the new equation: $$g = 4 \pi^{2} \frac{0.75000\; m}{(1.7357\; s)^{2}} \ldotp$$

Calculate to find g: $$g = 9.8281\; m/s^{2} \ldotp$$

Significance

This method for determining g can be very accurate, which is why length and period are given to five digits in this example. For the precision of the approximation sin $\theta$ ≈ $\theta$ to be better than the precision of the pendulum length and period, the maximum displacement angle should be kept below about 0.5°.

Exercise $\PageIndex{1}$

An engineer builds two simple pendulums. Both are suspended from small wires secured to the ceiling of a room. Each pendulum hovers 2 cm above the floor. Pendulum 1 has a bob with a mass of 10 kg. Pendulum 2 has a bob with a mass of 100 kg. Describe how the motion of the pendulums will differ if the bobs are both displaced by 12°.

Physical Pendulum

Any object can oscillate like a pendulum. Consider a coffee mug hanging on a hook in the pantry. If the mug gets knocked, it oscillates back and forth like a pendulum until the oscillations die out. We have described a simple pendulum as a point mass and a string. A physical pendulum is any object whose oscillations are similar to those of the simple pendulum, but cannot be modeled as a point mass on a string, and the mass distribution must be included into the equation of motion.

As for the simple pendulum, the restoring force of the physical pendulum is the force of gravity. With the simple pendulum, the force of gravity acts on the center of the pendulum bob. In the case of the physical pendulum, the force of gravity acts on the center of mass (CM) of an object. The object oscillates about a point O. Consider an object of a generic shape as shown in Figure $\PageIndex{2}$.

Figure $\PageIndex{2}$: A physical pendulum is any object that oscillates as a pendulum, but cannot be modeled as a point mass on a string. The force of gravity acts on the center of mass (CM) and provides the restoring force that causes the object to oscillate. The minus sign on the component of the weight that provides the restoring force is present because the force acts in the opposite direction of the increasing angle $\theta$.

When a physical pendulum is hanging from a point but is free to rotate, it rotates because of the torque applied at the CM, produced by the component of the object’s weight that acts tangent to the motion of the CM. Taking the counterclockwise direction to be positive, the component of the gravitational force that acts tangent to the motion is −mg sin $\theta$. The minus sign is the result of the restoring force acting in the opposite direction of the increasing angle. Recall that the torque is equal to $\vec{\tau} = \vec{r} \times \vec{F}$. The magnitude of the torque is equal to the length of the radius arm times the tangential component of the force applied, |$\tau$| = rFsin$\theta$. Here, the length L of the radius arm is the distance between the point of rotation and the CM. To analyze the motion, start with the net torque. Like the simple pendulum, consider only small angles so that sin $\theta$ ≈ $\theta$. Recall from Fixed-Axis Rotation on rotation that the net torque is equal to the moment of inertia I = $\int$r2 dm times the angular acceleration $\alpha$, where \(\alpha = \frac{d^{2} \theta}{dt^{2}}:

\[I \alpha = \tau_{net} = L (-mg) \sin \theta \ldotp\]

Using the small angle approximation and rearranging:

\[\begin{split} I \alpha & = -L (mg) \theta; \\ I \frac{d^{2} \theta}{dt^{2}} & = -L (mg) \theta; \\ \frac{d^{2} \theta}{dt^{2}} & = - \left(\dfrac{mgL}{I}\right) \theta \ldotp \end{split}\]

Once again, the equation says that the second time derivative of the position (in this case, the angle) equals minus a constant $\left(− \dfrac{mgL}{I}\right)$ times the position. The solution is

\[\theta (t) = \Theta \cos (\omega t + \phi),\]

where $\Theta$ is the maximum angular displacement. The angular frequency is

\[\omega = \sqrt{\frac{mgL}{I}} \ldotp \label{15.20}\]

The period is therefore

\[T = 2 \pi \sqrt{\frac{I}{mgL}} \ldotp \label{15.21}\]

Note that for a simple pendulum, the moment of inertia is I = $\int$r2dm = mL2 and the period reduces to T = 2$\pi \sqrt{\frac{L}{g}}$.

Example $\PageIndex{2}$: Reducing the Swaying of a Skyscraper

In extreme conditions, skyscrapers can sway up to two meters with a frequency of up to 20.00 Hz due to high winds or seismic activity. Several companies have developed physical pendulums that are placed on the top of the skyscrapers. As the skyscraper sways to the right, the pendulum swings to the left, reducing the sway. Assuming the oscillations have a frequency of 0.50 Hz, design a pendulum that consists of a long beam, of constant density, with a mass of 100 metric tons and a pivot point at one end of the beam. What should be the length of the beam?

Strategy

We are asked to find the length of the physical pendulum with a known mass. We first need to find the moment of inertia of the beam. We can then use the equation for the period of a physical pendulum to find the length.

Solution

Find the moment of inertia for the CM.

Use the parallel axis theorem to find the moment of inertia about the point of rotation: $$I = I_{CM} + \frac{L^{2}}{4} M = \frac{1}{12} ML^{2} + \frac{1}{4} ML^{2} = \frac{1}{3} ML^{2} \ldotp$$

The period of a physical pendulum has a period of T = 2$\pi \sqrt{\frac{I}{mgL}}$. Use the moment of inertia to solve for the length L: $$\begin{split} T & = 2 \pi \sqrt{\frac{I}{mgL}} = 2 \pi \sqrt{\frac{\frac{1}{3} ML^{2}}{MgL}} = 2 \pi \sqrt{\frac{L}{3g}}; \\ L & = 3g \left(\dfrac{T}{2 \pi}\right)^{2} = 3 (9.8\; m/s^{2}) \left(\dfrac{2\; s}{2 \pi}\right)^{2} = 2.98\; m \ldotp \end{split}$$

This length L is from the center of mass to the axis of rotation, which is half the length of the pendulum. Therefore the length H of the pendulum is: $$ H = 2L = 5.96 \: m $$

Significance

There are many ways to reduce the oscillations, including modifying the shape of the skyscrapers, using multiple physical pendulums, and using tuned-mass dampers.

Torsional Pendulum

A torsional pendulum consists of a rigid body suspended by a light wire or spring (Figure $\PageIndex{3}$). When the body is twisted some small maximum angle ($\Theta$) and released from rest, the body oscillates between ($\theta$ = + $\Theta$) and ($\theta$ = − $\Theta$). The restoring torque is supplied by the shearing of the string or wire.

Figure $\PageIndex{3}$: A torsional pendulum consists of a rigid body suspended by a string or wire. The rigid body oscillates between ($\theta$ = + $\Theta$) and ($\theta$ = − $\Theta$).

The restoring torque can be modeled as being proportional to the angle:

\[\tau = - \kappa \theta \ldotp\]

The variable kappa ($\kappa$) is known as the torsion constant of the wire or string. The minus sign shows that the restoring torque acts in the opposite direction to increasing angular displacement. The net torque is equal to the moment of inertia times the angular acceleration:

\[\begin{split} I \frac{d^{2} \theta}{dt^{2}} & = - \kappa \theta; \\ \frac{d^{2} \theta}{dt^{2}} & = - \frac{\kappa}{I} \theta \ldotp \end{split}\]

This equation says that the second time derivative of the position (in this case, the angle) equals a negative constant times the position. This looks very similar to the equation of motion for the SHM $\frac{d^{2} x}{dt^{2}}$ = − $\frac{k}{m}$x, where the period was found to be T = 2$\pi \sqrt{\frac{m}{k}}$. Therefore, the period of the torsional pendulum can be found using

\[T = 2 \pi \sqrt{\frac{I}{\kappa}} \ldotp \label{15.22}\]

The units for the torsion constant are [$\kappa$] = N • m = (kg • m/s2)m = kg • m2/s2 and the units for the moment of inertial are [I] = kg • m2, which show that the unit for the period is the second.

Example $\PageIndex{3}$: Measuring the Torsion Constant of a String

A rod has a length of l = 0.30 m and a mass of 4.00 kg. A string is attached to the CM of the rod and the system is hung from the ceiling (Figure $\PageIndex{4}$). The rod is displaced 10° from the equilibrium position and released from rest. The rod oscillates with a period of 0.5 s. What is the torsion constant $\kappa$?

Figure $\PageIndex{4}$ - (a) A rod suspended by a string from the ceiling. (b) Finding the rod’s moment of inertia.

Strategy

We are asked to find the torsion constant of the string. We first need to find the moment of inertia.

Solution

Find the moment of inertia for the CM: $$I_{CM} = \int x^{2} dm = \int_{- \frac{L}{2}}^{+ \frac{L}{2}} x^{2} \lambda dx = \lambda \Bigg[ \frac{x^{3}}{3} \Bigg]_{- \frac{L}{2}}^{+ \frac{L}{2}} = \lambda \frac{2L^{3}}{24} = \left(\dfrac{M}{L}\right) \frac{2L^{3}}{24} = \frac{1}{12} ML^{2} \ldotp$$

Calculate the torsion constant using the equation for the period: $$\begin{split} T & = 2 \pi \sqrt{\frac{I}{\kappa}}; \\ \kappa & = I \left(\dfrac{2 \pi}{T}\right)^{2} = \left(\dfrac{1}{12} ML^{2}\right) \left(\dfrac{2 \pi}{T}\right)^{2}; \\ & = \Big[ \frac{1}{12} (4.00\; kg)(0.30\; m)^{2} \Big] \left(\dfrac{2 \pi}{0.50\; s}\right)^{2} = 4.73\; N\; \cdotp m \ldotp \end{split}$$

Significance

Like the force constant of the system of a block and a spring, the larger the torsion constant, the shorter the period.

This page titled 15.5: Pendulums is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

15.4: Comparing Simple Harmonic Motion and Circular Motion

15.6: Damped Oscillations

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PENDULUM中文(繁體)翻譯：劍橋詞典

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pendulum 在英語-中文（繁體）詞典中的翻譯

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/ˈpen.dʒəl.əm/ us

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[ C ] a device consisting of a weight on a stick or thread that moves from one side to the other, especially one that forms a part of some types of clocks

擺，擺錘;（尤指）鐘擺

The pendulum in the grandfather clock swung back and forth.

落地鐘的鐘擺來回擺動。

The spaceship's jets were fired periodically to dampen a side-to-side pendulum motion that had developed.

太空船的噴射裝置週期性地噴射，以減緩太空船出現的左右搖擺現象。

[ S ] a change, especially from one opinion to an opposite one

搖擺不定的事態（或局面）;（尤指）觀點的搖擺不定

As so often in education, the pendulum has swung back to the other extreme and testing is popular again.

人們的觀點又回到了另一個極端，考試又流行起來了，這種搖擺不定的現象在教育界經常出現。

(pendulum在劍橋英語-中文（繁體）詞典的翻譯 © Cambridge University Press)

pendulum的例句

pendulum

Without knowing more about the particulars - for instance, about engineers'concepts of pendulums - one cannot dare to give a definitive answer in this case.

來自 Cambridge English Corpus

A natural question is whether (non-trivial) monodromy also can be defined for non-integrable perturbations of the spherical pendulum.

來自 Cambridge English Corpus

In particular, our approach applies to the spherical pendulum.

來自 Cambridge English Corpus

The entire traction drive is mounted on an assisting beam in the center of the bogie, and attached to the outer sides via two pendulums.

來自 Wikipedia

該例句來自維基百科，在CC BY-SA許可下可重複使用。

In these, a mechanical switch or a phototube turned an electromagnet on for a brief section of the pendulum's swing.

來自 Wikipedia

該例句來自維基百科，在CC BY-SA許可下可重複使用。

Consider a long line of pendulums, all pointing straight down.

來自 Wikipedia

該例句來自維基百科，在CC BY-SA許可下可重複使用。

All the pendulums are coupled together by the suspension.

來自 Cambridge English Corpus

In particular, he championed the use of pendulums for dowsing, also publishing arguments regarding the existence of ghosts and extraterrestrial involvement in human evolution.

來自 Wikipedia

該例句來自維基百科，在CC BY-SA許可下可重複使用。

示例中的觀點不代表劍橋詞典編輯、劍橋大學出版社和其許可證頒發者的觀點。

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a type of food similar to a hamburger but made without meat, by pressing together small pieces of vegetables, seeds, etc. into a flat, round shape

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Pendulums – The Physics Hypertextbook

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Discussion

the simple pendulum

A pendulum is a mass suspended from a pivot point that is free to swing back and forth. Because the motion is oscillatory (a fancy way to say back and forth) and periodic (repeating with a characteristic time), pendulums have been used in clocks since the 17th century. Crude pendulums are cheap and easy to build — all you need is a small weight, a piece of string, and something to hang it from — and make appropriate hands-on devices for introductory physics courses. The arms and legs of a walking person are also pendulums, so your typical off-the-shelf human comes with pendulums included.

A simple pendulum is a mathematical idealization used to approximate the behavior of real pendulums. Simple pendulums do not exist, but they're how real physics gets done. Start with the simple and build up to the real. All the mass of a simple pendulum is concentrated at a single point (called the bob) on the end of an unstretchable, incompressible, massless rod connected to a frictionless pivot that does not move. The bob of a simple pendulum traces back and forth over the arc of a circle with the pivot at the center. A stationary pendulum of any sort will rest in an equilibrium position with its center of mass below the pivot.

Two forces act on the bob a simple pendulum whether it is moving or not — weight (the force of gravity) and tension. (Air resistance is ignored, as is often the case. Poor air resistance. Always being ignored.) In the equilibrium position, when the pendulum is at rest, the two forces are equal and opposite. Not very interesting.

Since the whole pendulum is held fixed at the pivot (the noun), any attempt to move the pendulum side to side causes the whole thing to pivot (the verb) with an angular displacement θ. Now things get interesting.

This is a problem where a radial-tangential coordinate system works better than the typical vertical-horizontal one. Tension points radially inward toward the pivot and is considered the "good" vector. Weight is the "bad" vector and must be broken up into components — one radial (mg sin θ) and one tangential (mg cos θ).

The radial component and the tension contribute to the centripetal force. Tension is the positive one since it points toward the center. The radius (r) in the centripetal acceleration equation could be replaced with the length of the pendulum (ℓ) since they are the same thing. The tangential velocity (v) can also be written as a derivative for those who like calculus. There's no real reason to do that now other than because it looks cool. Actually, there's no reason to write any of this now other than for a sense of completeness.

∑Fr =

mac

T − mg cos θ =

⎛⎜⎝

⎞2⎟⎠

ℓ

The tangential component is the more interesting one. It's described a restoring force because it increases with increasing displacement and acts to push the bob back toward its equilibrium position. For that reason, it gets a negative sign. When displacement is positive, it points in the negative direction. When displacement is negative, it points in the positive direction. In informal language, it does the opposite. Now the tangential acceleration really should be written in fancy calculus notation. It's more useful that way. Also the mass cancels out, which is interesting.

∑Ft =

mat

− mg sin θ =

− g sin θ =

d2s

dt2

This is a second order differential equation that is extremely difficult to solve. I have never done it myself. You might think that's the end of this discussion, but you've forgotton one thing. This is a simple pendulum. It's already full of assumptions that aren't true (point masses, unstretchable strings, etc.). One more simplification couldn't hurt.

small angle approximation

Go back to the definition of sine as the ratio of the side opposite (x) the angle (θ) to the hypotenuse (r). Here's a (possibly new) fact for you known as the small angle approximation: the opposite side length (x) approaches the arc length (s) as the angle (θ) approaches zero (0).

θ → 0

⇒

x → s

This means that…

sin θ =

can be replaced with…

sin θ ≈

ℓ

…for a simple pendulum when the angle is small. That gives us a differential equation that is much easier to solve.

− g sin θ =

d2s

dt2

− g

ℓ

d2s

dt2

s =

dt2

−

ℓ

We need a function whose second derivative is itself with a minus sign. We have two options: sine and cosine. Either one is fine since they're basically identical functions with a 90° phase shift between them. Without loss of generality, I'll choose sine with an arbitrary phase angle (φ) that could equal 90° if we let it. Or it equal 0° or some other angle. The other parameters in a generic sine function are displacement amplitude (s0) and angular frequency (ω).

The basic method I've started is called "guess and check". My guess is that the function looks like a generic sine function…

s = s0 sin(ωt + φ)

and the check is to pop it back into the differential equation and see what happens.

s0 sin(ωt + φ)

= −

s0 sin(ωt + φ)

dt2

ℓ

− ω2s0 sin(ωt + φ)

= −

s0 sin(ωt + φ)

ℓ

ω2

ℓ

Basically everything cancels but one of my three parameters — angular frequency.

ω = √

ℓ

I don't think in angular frequencies. They're too abstract. I want the frequency — frequency with no adjective in front. Something that uses hertz [Hz] as the unit.

f =

√

2π

ℓ

Actually, I don't even want that. Pendulums tend to be slow beasts. I'd prefer the reciprocal of frequency — period. The period of a simple pendulum with small angle amplitudes is given by the following equation…

T = 2π√

ℓ

Where…

T =

period [s], the time to complete one cycle of motion

π =

a mathematical constant [unitless]

ℓ =

length [m], measured from the suspension point to the mass

g =

gravity [m/s2], the gravitational field strength in the place where the pendulum is doing its thing

What this says about simple pendulums…

The period of a simple pendulum is independent of it mass. Light bob, heavy bob, no difference. This should make sense since the acceleration of a freely falling body is independent of its mass. For practical reasons, however, mass may be worth considering when air resistance is added in. Heavy masses are less affected by aerodynamic drag than lighter ones since acceleration is inversely proportional to mass. Do you want your pendulum to keep plowing through the air? Then make it massive.

The period of a simple pendulum is proportional to the square root of its length (T ∝ √ℓ). Longer pendulums take longer to swing back and forth once than short ones. Doubling the period of a simple pendulum requires a quadrupling of its length. This should make intuitive sense on some level. Long legs take longer to swing back and forth than short legs do. A elephant takes fewer footsteps than a mouse would at the same speed.

The period of a simple pendulum is inversely proportional to the square root of the local gravitational field strength (T ∝ 1/√g). Doesn't quite roll off the tongue nicely, does it? Quadrupling gravity would halve the period of a simple pendulum. Not really a situation I could see myself encountering. Local gravity doesn't vary much across the surface of the Earth — at least as far as direct human perception is concerned — but it matters when pendulums are how you keep accurate track of time.

What this doesn't say about simple pendulums…

The period of a simple pendulum increases with increasing amplitude in a way that is difficult to describe, difficult to explain, and frequently unimportant.

large angle correction

The equation for the period of a simple pendulum that accounts for large angles is the small angle approximation…

T = 2π√

ℓ

multiplied by an infinite sum of ever smaller corrections. The first 5 terms are shown below. I will not attempt to derive this equation.

T = 2π√

ℓ

⎡⎢⎣

1 +

⎛⎜⎝

⎞2⎟⎠

sin2

⎛⎜⎝

⎞⎟⎠

⎛⎜⎝

1 · 3

⎞2⎟⎠

sin4

⎛⎜⎝

⎞⎟⎠

2 · 4

⎛⎜⎝

1 · 3 · 5

⎞2⎟⎠

sin6

⎛⎜⎝

⎞⎟⎠

2 · 4 · 6

⎛⎜⎝

1 · 3 · 5 · 7

⎞2⎟⎠

sin8

⎛⎜⎝

⎞⎟⎠

+ …

⎤⎥⎦

2 · 4 · 6 · 8

Since every term in this infinite beast is positive, the true period of a simple pendulum will always be greater than that calculated using the small angle approximation. The correction adds <0.1% up to 7°, <1% up to 22°, <10% up to 69°, and about 18% at 90°.

If even larger angles are allowed, then the large angle correction maxes out at +116% for pendulums with an angle of release of 180°.

Another version of the infinite sum of corrections is shown below. This one starts with the equation derived above and replaces all the sine functions with their Taylor expansions (each one an infinite sum unto itself). Collect like terms, remove common factors, and this is what you get…

T = 2π√

ℓ

⎡⎢⎣

1 +

θ2

θ4

3,072

173

θ6

737,280

22,931

θ8

+ …

⎤⎥⎦

1,321,205,760

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11.3: Pendulums

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Julio Gea-BanaclocheUniversity of Arkansas via University of Arkansas Libraries

The Simple PendulumThe "Physical Pendulum"

The Simple Pendulum

Besides masses on springs, pendulums are another example of a system that will exhibit simple harmonic motion, at least approximately, as long as the amplitude of the oscillations is small. The simple pendulum is just a mass (or “bob”), approximated here as a point particle, suspended from a massless, inextensible string, as in Figure $\PageIndex{1}$.

We could analyze the motion of the bob by using the general methods introduced in Chapter 8 to deal with motion in two dimensions—breaking down all the forces into components and applying $\vec{F}_{net} = m\vec{a}$ along two orthogonal directions—but this turns out to be complicated by the fact that both the direction of motion and the direction of the acceleration are constantly changing. Although, under the assumption of small oscillations, it turns out that simply using the vertical and horizontal directions is good enough, this is not immediately obvious, and arguably it is not the most pedagogical way to proceed.

Figure $\PageIndex{1}$: A simple pendulum. The mass of the bob is $m$, the length of the string is $l$, and torques are calculated around the point of suspension O. The counterclockwise direction is taken as positive.

Instead, I will take advantage of the obvious fact that the bob moves on an arc of a circle, and that we have developed already in Chapter 9 a whole set of tools to deal with that kind of motion. Let us, therefore, describe the position of the pendulum by the angle it makes with the vertical, $\theta$, and let $\alpha = d^2 \theta /dt^2$ be the angular acceleration; we can then write the equation of motion in the form $\tau_{net} = I\alpha$, with the torques taken around the center of rotation—which is to say, the point from which the pendulum is suspended. Then the torque due to the tension in the string is zero (since its line of action goes through the center of rotation), and $\tau_{net}$ is just the torque due to gravity, which can be written

\[ \tau_{n e t}=-m g l \sin \theta \label{eq:11.19} .\]

The minus sign is there to enforce a consistent sign convention for $\theta$ and $\tau$: if, for instance, I choose counterclockwise as positive for both, then I note that when $\theta$ is positive (pendulum to the right of the vertical), $\tau$ is clockwise, and hence negative, and vice-versa. This is characteristic of a restoring torque, that is to say, one that will always try to push the system back to its equilibrium position (the vertical in this case).

As for the moment of inertia of the bob, it is just $I = ml^2$ (if we treat it as just a point particle), so the equation $\tau_{net} = I\alpha$ takes the form

\[ m l^{2} \frac{d^{2} \theta}{d t^{2}}=-m g l \sin \theta \label{eq:11.20} .\]

The mass and one factor of $l$ cancel, and we get

\[ \frac{d^{2} \theta}{d t^{2}}=-\frac{g}{l} \sin \theta \label{eq:11.21} .\]

Equation (\ref{eq:11.21}) is an example of what is known as a differential equation. The problem is to find a function of time, $\theta (t)$, that satisfies this equation; that is to say, when you take its second derivative the result is equal to $−(g/l) \sin[\theta (t)]$. Such functions exist and are called elliptic functions; they are included in many modern mathematical packages, but they are still not easy to use. More importantly, the oscillations they describe, in general, are not of the simple harmonic type.

On the other hand, if the amplitude of the oscillations is small, so that the angle $\theta$, expressed in radians, is a small number, we can make an approximation that greatly simplifies the problem, namely,

\[ \sin \theta \simeq \theta \label{eq:11.22} .\]

This is known as the small angle approximation, and requires $\theta$ to be in radians. As an example, if $\theta$ = 0.2 rad (which corresponds to about 11.5$^{\circ}$), we find $\sin \theta$ = 0.199, to three-figure accuracy.

With this approximation, the equation to solve become much simpler:

\[ \frac{d^{2} \theta}{d t^{2}}=-\frac{g}{l} \theta \label{eq:11.23} .\]

We have, in fact, already solved an equation completely equivalent to this one in the previous section: that was equation (11.2.8) for the mass-on-a-spring system, which can be rewritten as

\[ \frac{d^{2} x}{d t^{2}}=-\frac{k}{m} x \label{eq:11.24} \]

since $a = d^2x/dt^2$. Just like the solutions to (\ref{eq:11.24}) could be written in the form $x(t) = A \cos(\omega t + \phi)$, with $\omega=\sqrt{k / l}$, the solutions to (\ref{eq:11.23}) can be written as

\begin{align}

\theta(t) &=A \cos (\omega t+\phi) \nonumber \\

\omega &=\sqrt{\frac{g}{l}} \label{eq:11.25}.

\end{align}

This tells us that if a pendulum is not pulled too far away from the vertical (say, about 10$^{\circ}$ or less) it will perform approximate simple harmonic oscillations, with a period of

\[ T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{l}{g}} \label{eq:11.26} .\]

This depends only on the length of the pendulum, and remains constant even as the oscillations wind down, which is why it became the basis for time-keeping devices, beginning with the invention of the pendulum clock by Christiaan Huygens in 1656. In particular, a pendulum of length $l$ = 1 m will have a period of almost exactly 2 s, which is what gives you the familiar “tick-tock” rhythm of a “grandfather’s clock,” once per second (that is to say, once every half period).

The "Physical Pendulum"

By a “physical pendulum” one means typically any pendulum-like device for which the moment of inertia is not given by the simple expression $I = ml^2$. This means that the mass is not concentrated into a single point-like particle a distance $l$ away from the point of suspension; rather, for example, the bob could have a size that is not negligible compared to $l$ (as in Figure 11.1.1), or the “string” could have a substantial mass of its own—it could, for instance, be a chain, like in a playground swing, or a metal rod, as in most pendulum clocks.

Figure (b) shows the special case of a thin rod of length $l$ pivoted at one end (the distance $d = l/2$ in this case). In both cases, there is an additional force (not shown) acting at the pivot point, to balance gravity.

Regardless of the reason, having to deal with a distributed mass means also that one needs to use the center of mass of the system as the point of application of the force of gravity. When this is done, the motion of the pendulum can again be described by the angle between the vertical and a line connecting the point of suspension and the center of mass. If the distance between these two points is $d$, then the torque due to gravity is $−mgd \sin \theta$, and the only other force on the system, the force at the pivot point, exerts no torque around that point, so we can write the equation of motion in the form

\[ I \frac{d^{2} \theta}{d t^{2}}=-m g d \sin \theta \label{eq:11.27} .\]

Under the small-angle approximation, this will again lead to simple harmonic motion, only now with an angular frequency given by

\[ \omega=\sqrt{\frac{m g d}{I}} \label{eq:11.28} .\]

As an example, consider the oscillations of a uniform, thin rod of length $l$ and mass $m$ pivoted at one end. We then have $I = ml^{2}/3$, and $d = l/2$, so Equation (\ref{eq:11.28}) gives

\[ \omega=\sqrt{\frac{3 g}{2 l}} \label{eq:11.29} .\]

This is about 22% larger than the result (\ref{eq:11.25}) for a simple pendulum of the same length, implying a correspondingly shorter period.

This page titled 11.3: Pendulums is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Julio Gea-Banacloche (University of Arkansas Libraries) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.

11.2: Simple Harmonic Motion

11.4: In Summary

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PENDULUM | English meaning - Cambridge Dictionary

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English

Meaning of pendulum in English

pendulumnoun uk

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/ˈpen.dʒəl.əm/ us

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/ˈpen.dʒəl.əm/

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[ C ] a device consisting of a weight on a stick or thread that moves from one side to the other, especially one that forms a part of some types of clocks: The pendulum in the grandfather clock swung back and forth. The spaceship's jets were fired periodically to dampen a side-to-side pendulum motion that had developed.

TATSUHIKO SAWADA/amana images/GettyImages

[ S ] a change, especially from one opinion to an opposite one: As so often in education, the pendulum has swung back to the other extreme and testing is popular again.

SMART Vocabulary: related words and phrases

Watches & clocks

24-hour clock

against the clock idiom

alarm

alarm clock

at/on the stroke of something idiom

atomic clock

clock

horology

hour hand

hourglass

lose

military time

minute hand

movement

stroke

timepiece

turn the clocks back phrase

twenty-four-hour clock

unsynchronized

watchband

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You can also find related words, phrases, and synonyms in the topics:

Changing your mind

(Definition of pendulum from the Cambridge Advanced Learner's Dictionary & Thesaurus © Cambridge University Press)

pendulum | American Dictionary

pendulumnoun [ C ] us

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/ˈpen·dʒə·ləm/

Add to word list

a device consisting of a weight hanging on a rod or cord that moves from one side to the other, esp. one that forms a part of a clock

A pendulum is also power or control of an activity that changes from one group to another: In labor-management relations, the pendulum has swung wildly in the direction of the players.

(Definition of pendulum from the Cambridge Academic Content Dictionary © Cambridge University Press)

Examples of pendulum

pendulum

Despite some unavoidable differences between the theoretical and actual pendulums, the phenomena exhibited by them show good correspondence.