Two buckets, each of mass M, are attached to a light and elastic string with elastic constant k over a light and smooth pulley as follows:
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Nov09/CrazyQ1.jpg
The buckets have their base parts levelled and kept stationary.
Now, a small particle of mass m is dropped at a height h above the base of either bucket, as shown above.
Suppose that, at time t = 0, the particle reaches the base of the bucket, sticks to it, then moves together with the system as shown below:
圖片參考：http://i117.photobucket.com/albums/o61/billy_hywung/Nov09/CrazyQ2.jpg
Find, in terms of all given and/or other parameters, the expression of displacements x1 (Of the lighter bucket) and x2 (Of the heavier bucket) as functions of t. (Suppose that the string remains in tension all the time.)
Oscillating buckets
2009-11-21 1:08 am
回答 (4)
2009-12-04 4:20 pm
✔ 最佳答案
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參考資料：
ilovemadonna2009的提示 + my wisdom of physics + my wisdom of maths
2009-12-04 08:26:45 補充：
吓，ilovemadonna2009，乜原來kinetic energy和elastic potential energy的公式分別是1/2mv和1/2kx？？？
2009-12-04 08:45:00 補充：
據我所知一般system of differential equations的解法是把等號兩邊進行微分，但其實是否亦可以把等號兩邊進行積分？
如果是可以的話，這題仍有一線生機化成x1或x2獨立存在的微分方程，否則就真是……
2009-12-04 09:08:29 補充：
雖然我不敢保證該system of differential equations是100%可解，但是，我也可以估計到該兩個桶的運動模式是怎樣。
首先，該兩個桶肯定會各自上下擺動，但是由於該兩個桶的重量不同，較重的那個桶會傾向於向下移動而較輕的那個桶會傾向於向上移動，兩者合併後的的運動模式就會好像y = x + sin x的那樣。
但是由於elastic string的長度始終有限，因此較輕的那個桶最終會被滑輪頂住，到時較重的那個桶仍會不斷上下擺動，但是已再沒有傾向於向下移動的趨勢了。
2009-12-04 15:07:47 補充：
飛天魏國大將軍張遼，我並不是質疑你所出的問題是否有明確的答案，因為我與你一樣，亦是一個對科學非常開通的人，我會接受以no-close-form integral來表示答案，而且我亦會接受以unsolvable的differential equation來表示答案（注意，前者仍算是solvable，因為一條differential equation只要能夠找到完全沒有微分符號的expression就已經算是是solvable。）。
2009-12-04 15:24:47 補充：
例如large angle pendulum motion in linear-drag environment的differential equation，經過某些代入法降階後就會變成Abel equation of the second kind in the canonical form (http://eqworld.ipmnet.ru/en/solutions/ode/ode0124.pdf)（不信的話請你自己做下）。我會欣然接受去到Abel equation of the second kind in the canonical form的那一步就立即當是答案。
2009-12-04 15:35:38 補充：
回到這題，由於在這個system of differential equations內第一條differential equation的形式是較為簡單但較為高階，而第二條differential equation的形式是較為複雜但較為低階，再者把第二條differential equation先動手亦會變得更複雜，因此我打算把第一條differential equation先動手。但是這樣會出現007號意見所示的那條問題。希望有人對那條問題有正確的回覆。
2009-12-04 5:32 pm
換言之, 即是個 centre of mass of the system 會持續下降, 直到較輕的那個桶被滑輪頂住為止.
向一眾用家聲明, 我問的問題未必會有一個明確的答案, 惟過程步驟最為關鍵, 言亦是我比分的準則.
向一眾用家聲明, 我問的問題未必會有一個明確的答案, 惟過程步驟最為關鍵, 言亦是我比分的準則.
2009-12-04 4:26 am
Loss of G.P.E. = Gain in K.E. + Gain in E.P.E.
(M + m)gx2 - Mgx1 = 1/2 (M + m)x2' + 1/2 Mx1' + 1/2 k(x2 - x1)
[2(M + m)g - k]x2 - (2Mg - k)x1 = (M + m)x2' + Mx1' ... (3)
(M + m)gx2 - Mgx1 = 1/2 (M + m)x2' + 1/2 Mx1' + 1/2 k(x2 - x1)
[2(M + m)g - k]x2 - (2Mg - k)x1 = (M + m)x2' + Mx1' ... (3)
2009-11-21 4:39 am
Velocity of the particle just before hitting the bucket at t = 0, u = sqrt(2gh)
By conservation of momentum
mu = (M + m)v
Speed of the heavier bucket just after the impact, v = msqrt(2gh) /(M + m)
2009-11-20 20:40:40 補充：
Let T be the tension in the string.
Let L and e be the natural length and the extension of one side of the string initially.
By Hooke's Law, Mg = ke
Now, after the impact,
(M + m)g - T = (M + m)x2"
(M + m)g - k(2e + x2 - x1) = (M + m)x2"
mg - k(e + x2 - x1) = (M + m)x2" ... (1)
2009-11-20 20:40:52 補充：
T - Mg = Mx1"
k(2e + x2 - x1) - Mg = Mx1"
k(e + x2 - x1) = Mx1" ... (2)
By conservation of energy,
Loss of G.P.E. = Gain in K.E. + Gain in E.P.E.
(M + m)gx2 - Mgx1 = 1/2 (M + m)x2' + 1/2 Mx1' + 1/2 k(x2 - x1)
[2(M + m)g - k]x2 - (2Mg - k)x1 = (M + m)x2' + Mx1' ... (3)
By conservation of momentum
mu = (M + m)v
Speed of the heavier bucket just after the impact, v = msqrt(2gh) /(M + m)
2009-11-20 20:40:40 補充：
Let T be the tension in the string.
Let L and e be the natural length and the extension of one side of the string initially.
By Hooke's Law, Mg = ke
Now, after the impact,
(M + m)g - T = (M + m)x2"
(M + m)g - k(2e + x2 - x1) = (M + m)x2"
mg - k(e + x2 - x1) = (M + m)x2" ... (1)
2009-11-20 20:40:52 補充：
T - Mg = Mx1"
k(2e + x2 - x1) - Mg = Mx1"
k(e + x2 - x1) = Mx1" ... (2)
By conservation of energy,
Loss of G.P.E. = Gain in K.E. + Gain in E.P.E.
(M + m)gx2 - Mgx1 = 1/2 (M + m)x2' + 1/2 Mx1' + 1/2 k(x2 - x1)
[2(M + m)g - k]x2 - (2Mg - k)x1 = (M + m)x2' + Mx1' ... (3)
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