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來投個文章試試||傅科擺問題的兩種方法分析

2021-01-07 09:09 作者:湮滅的末影狐  | 我要投稿

//現(xiàn)在是2021.1.7,閑著無聊,并發(fā)現(xiàn)B站專欄文章居然支持LaTeX公式系統(tǒng)。

//所以隨便寫點之前的東西來玩玩。

//感謝Mathpix的公式識別!我真的不會想用TeX再把那些公式打一遍...

//這是去年物理研討課程中我的報告內(nèi)容。由于老師鼓勵用報告用英文寫,這里是直接將之前的報告原文搬運過來的,所以是全英文...請原諒英語渣的語法錯誤...

Abstract

This article demonstrates?two different methods to solve the problem with the example of Foucault pendulum. We will find the exact motion of the pendulum by using both algebraic and geometric method and evaluate the advantages and disadvantages of the two methods. And we will discuss about how we choose our method to solve a new problem, and how our geometrical intuition help us to make a breakthrough.

1. Introduction

A Foucault Pendulum is a large mass suspended from a long line, which is often used as an example for the effect of Coriolis force. When it swings in a vertical plain, the earth rotates beneath it, creating a relative motion between them so that we can see the perpendicular plain of swing rotates slowly. In this article we will also discuss general Foucault Pendulum, where we put the pendulum on a?rotational reference frame with any angular velocity.

Figure1. A Foucault Pendulum

We will?find out the angular frequency ??′ of the perpendicular plain’s rotation.

2. Algebraic Method

In this part we analyze the tragectory of the pendulum by solving the differential eqation of the pendulum's motion. (原諒畫渣的全損畫質(zhì)示意圖)


FIgure2. Approximate analyze of force

For a simple pendulum with mass ?? and length of the string ??, we know that the net force on the mass is

????????????????????????????????????????????????????????????%5Cvec%20F(%5Cvec%20r)%3D-m%20%5Comega_0%5E2%20%5Cvec%20r????????????????????????????????????????????????????(1)

Where?%5Comega_0%3D%5Csqrt%7B%5Cfrac%20gl%7D?is the original angular frequency of the pendulum. %5Cvec%20r%3D(x%2Cy)?is the horizontal positon vector.

Figure3 Demonstration of pendulum

We assume that the Earth rotates with angular frequency? %5Comega_e. Consider the effect of inertial centrifugal force, the inertial force is

???????????????????????????????????%5Cvec%7BF%7D_%7Bi%7D%3D2%20m%20%5Cvec%7Bv%7D%20%5Ctimes%20%5Cvec%7B%5Comega%7D_%7Be%7D-m%20%5Cvec%7B%5Comega%7D_%7Be%7D%20%5Ctimes%5Cleft(%5Cvec%7B%5Comega%7D_%7Be%7D%20%5Ctimes%20%5CDelta%20%5Cvec%7Br%7D%5Cright)????????????????????????????????????(2)

where?%5CDelta%20%5Cvec%20r%20%3D%20(x%2Cy%2C0) because we neglect the vertical?motion (that is z direction).

And we can find that

??????????????????????????????????????F_%7Bi%20x%7D%3D2%20m%20%5Cdot%7By%7D%20%5Comega_%7Be%7D%20%5Csin%20%5Ctheta%2Bm%20%5Comega_%7Be%7D%5E%7B2%7D%20%5Csin%20%5E%7B2%7D%20%5Ctheta%20%5Ccdot%20x????????????????????????????????????(3)

????????????????????????????????????F_%7Bi%20y%7D%3D-2%20m%20%5Cdot%7Bx%7D%20%5Comega_%7Be%7D%20%5Csin%20%5Ctheta%2Bm%20%5Comega_%7Be%7D%5E%7B2%7D%20%5Csin%20%5E%7B2%7D%20%5Ctheta%20%5Ccdot%20y???????????????????????????????????(4)

where?%5Ctheta is the latitude of the lab.

Then we set the equations of horizontal motion up by using Newton’s second law:?

????????????????????????????????????%5Cddot%7Bx%7D%3D-%5Comega_%7B0%7D%5E%7B2%7D%20x%2B2%20%5Cdot%7By%7D%20%5Comega_%7Be%7D%20%5Csin%20%5Ctheta%2B%5Comega_%7Be%7D%5E%7B2%7D%20%5Csin%20%5E%7B2%7D%20%5Ctheta%20%5Ccdot%20x???????????????????????????????? (5)

????????????????????????????????????%5Cddot%7By%7D%3D-%5Comega_%7B0%7D%5E%7B2%7D%20y-2%20%5Cdot%7Bx%7D%20%5Comega_%7Be%7D%20%5Csin%20%5Ctheta%2B%5Comega_%7Be%7D%5E%7B2%7D%20%5Csin%20%5E%7B2%7D%20%5Ctheta%20%5Ccdot%20y??????????????????????????????????(6)

This is a linear system. And if we let complex position?%5Ctilde%20s%20%3D%20x%2Bi%20y, and let?(5)%2B(6)%5Ccdot%20i, we will find that the two equations can be written as

?????????????????????????%5Cfrac%7Bd%5E%7B2%7D%20%5Ctilde%7Bs%7D%7D%7Bd%20t%5E%7B2%7D%7D%2B2%20i%20%5Comega_%7Be%7D%20%5Csin%20%5Ctheta%20%5Ccdot%20%5Cfrac%7Bd%20%5Ctilde%7Bs%7D%7D%7Bd%20t%7D%2B%5Cleft(%5Comega_%7B0%7D%5E%7B2%7D-%5Comega_%7Be%7D%5E%7B2%7D%20%5Csin%20%5E%7B2%7D%20%5Ctheta%5Cright)%20%5Ctilde%7Bs%7D%3D0???????????????????????? (7)

and there's only one variable in this complex-variable differential equation. (That's quite funny as we try to use "complex" to "simplify" our question)

Figure 4. Use complex number to simplify calculation

The equation has a general solution:

???????????????????????????????????????????????????????%5Ctilde%20s%20%3D%5Ctilde%20C_1%20e%5E%7B%5Clambda_1%20t%7D%2B%20%5Ctilde%20C_2%20e%5E%7B%5Clambda_2%20t%7D???????????????????????????????????????????????(8)

Where %5Ctilde%20C_1%2C%5Ctilde%20C_2?are complex const determined by initial motion of the pendulum, and

????????????????????????????????????????????????????%5Clambda_%7B1%2C2%7D%3D-i%20%5Comega_%7Be%7D%20%5Csin%20%5Ctheta%20%5Cpm%20i%20%5Comega_%7B0%7D???????????????????????????????????????????? (9)

are the two roots of characteristic equation

????????????????????????????????????%5Clambda%5E%7B2%7D%2B2%20i%20%5Comega_%7Be%7D%20%5Csin%20%5Ctheta%20%5Ccdot%20%5Clambda%2B%5Comega_%7B0%7D%5E%7B2%7D-%5Comega_%7Be%7D%5E%7B2%7D%20%5Csin%20%5E%7B2%7D%20%5Ctheta%3D0????????????????????????????(10)

or the solution can be written as

??????????????????????????????????????????????????????%5Ctilde%20s%20%3D%20%5Ctilde%20C_1%20e%5E%20%7Bi%5Comega_1%20t%7D%2B%5Ctilde%20C_2%20e%5E%20%7Bi%5Comega_2%20t%7D????????????????????????????????????????????(11)

where

?????????????????????????????????????????????????????????%5Comega_%7B1%7D%3D%5Comega_%7B0%7D-%5Comega_%7Be%7D%20%5Csin%20%5Ctheta????????????????????????????????????????????????(12)

?????????????????????????????????????????????????????????%5Comega_%7B2%7D%3D%5Comega_%7B0%7D%2B%5Comega_%7Be%7D%20%5Csin%20%5Ctheta????????????????????????????????????????????????(13)

so that

????????????????????????????????????????%5Ctilde%7Bs%7D%20e%5E%7Bi%20%5Comega_%7Be%7D%20%5Csin%20%5Ctheta%20%5Ccdot%20t%7D%3D%5Ctilde%7BC%7D_%7B1%7D%20e%5E%7Bi%20%5Comega_%7B0%7D%20t%7D%2B%5Ctilde%7BC%7D_%7B2%7D%20e%5E%7B-i%20%5Comega_%7B0%7D%20t%7D????????????????????????????????????? (14)

Figure 5. Simulation of trajectory

which means that in a rotational reference frame with angular frequency %F0%9D%9C%94_%F0%9D%91%92%20%20%5Csin%E2%81%A1%F0%9D%9C%83, the pendulum looks just like a normal pendulum with angular frequency %5Comega_0. (寫到這里突然發(fā)現(xiàn)這里字數(shù)統(tǒng)計是按照字母算的...難怪這么快就2k字了)

So the angular frequency we’ve been trying to find is

??????????????????????????????????????????????????????????????%5Comega'%3D%5Comega_0%20%5Csin%20%5Ctheta????????????????????????????????????????????????????? (15)

3. Intuitionistic Geometrical Method

(公式轟炸終于結(jié)束了...如果有人看到這里了就給點耐心看下去吧,后面是比較輕松愉快的幾何分析...現(xiàn)在其實我自己也覺得前面公式放多了...)

Let’s first consider a Foucault pendulum placed at the Arctic pole,?and observe it from the space. The Earth rotates but there’s no force to turn the pendulum together. Obviously the angular frequency of the rotation is %F0%9D%9C%94_%F0%9D%91%92.

Figure 5. Picture of our parallel

And if we put it somewhere else, it's just like the earth remains still but we're moving along a parallel line continuously. Obviously we need to turn to keep us on a certain latitude, and that's why the perpendicular plain seems to rotate. So we need to find how fast we turn when we move a circle along a parallel?everyday.?We only need to find out the total angle ?? we turned this day.

For a polygon on the surface of a sphere with solid angle Ω and exterior angles %F0%9D%9B%BC_1%2C%F0%9D%9B%BC_2%2C%F0%9D%9B%BC_3%2C%E2%80%A6%F0%9D%9B%BC_%F0%9D%91%9B, its solid angle writes(事實上我并不知道這個定理怎么證)

????????????????????????????????????????????? ? ? ? ??????%5COmega%3D2%5Cpi-%5CSigma_%7Bk%3D1%7D%5En%20%5Calpha_k????????????????????????????????????????????????(16)

And %E2%88%91_%7B%F0%9D%91%98%3D1%7D%5E%F0%9D%91%9B%F0%9D%9B%BC_%F0%9D%91%98? is actually the total angle we turn when we walk along this polygon. That will also be true when we move along the parallel, and the solid angle determined by this parallel where we stand is (這里真的是最后的計算了)

?????????????????????????????%5COmega%3D%5Cfrac%7B1%7D%7Br%5E%7B2%7D%7D%20%5Cint_%7B0%7D%5E%7B%5Cfrac%7B%5Cpi%7D%7B2%7D-%5Ctheta%7D%202%20%5Cpi%20r%20%5Csin%20%5Cvarphi%20%5Ccdot%20r%20d%20%5Cvarphi%3D2%20%5Cpi(1-%5Csin%20%5Ctheta)????????????????????? (17)

so the angle we turn everyday will be

?????????????????????????????????????????????????%5Calpha%3D%5Clim%20_%7Bn%20%5Crightarrow%20%5Cinfty%7D%20%5Csum_%7Bk%3D1%7D%5E%7Bn%7D%20%5Calpha_%7Bk%7D%3D2%20%5Cpi%20%5Csin%20%5Ctheta????????????????????????????????????????(18)

But we can find a much easier way.

Figure 6. Create a cone to solve

Suppose a cone is tangent to the sphere and the tangent line is the parallel where we put the pendulum. And moving on the parallel of the sphere is equal to?moving?on the cone.?

Figure 7. Stretch out view of the cone surface

And we can easily find that?%5Calpha%3D2%5Cpi%5Csin%5Ctheta.

???????????????????????????????????????????????????????%5Comega'%3D%5Cfrac%7B%5Calpha%7D%7BT_e%7D%3D%5Comega_e%20%5Csin%20%5Ctheta????????????????????????????????????????????????(19)

That's how we use a completely geometrical method to solve the problem.

4. Conclusion

Both methods can solve the question correctly, but geometrical method is obviously easier. (寫到這里因為后面都是文本了就換了手機繼續(xù)寫...結(jié)果發(fā)現(xiàn)手機端好多格式都亂了...)

Nowadays we can easily solve differential equations with the help of computers, so geometrical method seems less useful than before. However the geometrical method brings a new way of thinking, to find another view?of the problem from space. Here we break the boundary of 2-dimensional space, and find a better solution of this problem. And if we break the boundary between space and time, we'll make a great breakthrough, theory of relativity. And that's why we need different views of a problem.

References

[1]辛國君,劉樹新,舒幼生.傅科擺的進動與軌跡的周期性[J].大學物理, 2013,32(04):5-7.

Acknowledgement

感謝我的室友們幫助審稿及討論,感謝李教授的指導(dǎo)與同學們的寶貴意見。

現(xiàn)在再去看又還是覺得有不少地方寫得不太好...畢竟半學期真的能學很多東西。現(xiàn)在會了Mathematica之類的一些新東西之后再看前面做的過程,其實很多困難是可以輕松處理的。不管怎樣這篇文章只是留作紀念,將來還有很長的路...

(當初準備這篇報告的時候挺有意思的就是...我找了一篇舒幼生教授的文章參考,并發(fā)現(xiàn)這篇文章里面的計算過程丟了一項離心力的變化量,導(dǎo)致計算結(jié)果的特征頻率與我的結(jié)果出現(xiàn)了一些偏差。在地球的低角速度下,以本題近似級別這個偏差是一個高階小量,但是如果在一個角速度無法看作小量的轉(zhuǎn)動系這一偏差會有點大,所以我堅持我的結(jié)果。)

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