Re: [題目] 普通物理震盪
※ 引述《sheepyPie (小羊派)》之銘言：
: [領域] 普通物理 (題目相關領域)
: [來源] 哈樂迪8e 15-37(d) (課本習題、考古題、參考書...)
: [題目] 15-37(D)
: 題目大概是有一彈簧(無質量,k)掛在天花板上下面放質量為0.4kg物體
: 平衡後的位置距離彈簧原本長度多遠?
: 補個圖
: ----------------------------------------------------
: │ │ │
: 彈 彈 彈
: 簧 簧 簧
: │ │ │
: 原長 ─ ● │ ─
: E_1 │ ↑
: │ ↓ y
: ● ─
: E_2
: [瓶頸] (寫寫自己的想法，方便大家為你解答)
: 我的想法是用能量守恆
: 在原長時 E_1 = 0
: 到最低點時 E_2 = -mgy + 1/2 k y^2
: 但是跟另外一個想法算出來的y不一樣
: 另外一個想法是 最下面靜力平衡
: F = ky = mg
: 第一個想法的答案是 2mg/k
: 第二個想法的答案是 mg/k
: 不知道那裡想錯了><
: 謝謝
let me try to explain this if this make sense to you,
first of all, i agree with your conservation of energy. Total energy in
the universe is niether created nor destroied.
But, I think you are mistaken the difference between
"conservation of energy" and "Energy difference"
In this case, you can set your "zero point" any place you want
However, if you are saying enegy at E_1 is zero, then the energy at E_2
"relative to" E_1 is the energy difference!!!! so particle's potential energy
at E_1 "does not" equal to the energy at E_2. Therefore, if you are trying
to calculate the "y", the way you did with Fs=Fg is correct, which gives
y=mg/k
___________________________________________________________
Aside: If you are saying "Conservation of energy", this could be an example
for your case. Assume that the same condition, if you release the mass from
position E_1, as it travels distance "y", then the potential energy released
(U=mgy) equal to the kinetic energy at position y away from the original
position with velocity "v"(T=1/2mv^2) PLUS the energy that stores inside the
spring (ES=1/2ky^2)
To clarify conservation of energy equation
→ U=T+ES
→mgy=1/2(mv^2+ky^2)
if you wish to find the velocity at that point
you can substitute y=mg/k into the equation and you can solve for v.X
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