Questions about SHM

2010-12-23 4:13 pm

圖片參考：http://imgcld.yimg.com/8/n/HA00414085/o/701012230015513873430440.jpg

Tension = ma = m(-w^2)x
Since m and w are constant, the tension is directly proportional to the displacement of oscillation.
But why the statement (2) is not true?

Here is another question.
If the direction of the tension is taken to be positive when it is pointing upward, is it true that the tension becomes negative when the bob is above the equlibrium position?

更新1:

我一開始就誤以為它是oscillating vertically.. 可不可以解答一下我的問題如果它是oscillating vertically?

回答 (2)

天同

2010-12-23 5:20 pm

✔ 最佳答案

The tension in the string T is given by,

T - mg.cos(theta) = mL.w^2
where m is the mass of the bob,
g is the acceleration due to gravity
(theta) is the angular displacement of the bob about the vertical
L is the length of the string
w is the angular speed of the bob

Hence, T = mg.cos(theta) + mL.w^2
since (theta) = A/L, where A is the amplitude of oscillation, thus when the bob is at an extreme position, w = 0 s^-1, T is proportional to cos(A/L). Clearly T is not proportional to A.

The direction of tension in the string is always pointing upward along the string throughout the oscillation. If not, it would not be able to balance the weight of the bob. There is no way that the string tension is pointing downward.

👍 0 👎 0

Atheist

2010-12-24 4:11 am

我一開始就誤以為它是oscillating vertically..
可不可以解答一下我的問題如果它是oscillating vertically?

👍 0 👎 0

收錄日期: 2021-04-29 17:37:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101223000051KK00155

檢視 Wayback Machine 備份