燒繩子估時間問題 - 推理遊戲
![Odelette avatar](/img/boy1.jpg)
By Odelette
at 2008-10-30T10:08
at 2008-10-30T10:08
Table of Contents
回一下鐘擺定理的解法
鐘擺週期公式是:T=2*pi*sqrt(L/g)
重點是,在擺幅不大、繩子長短固定的條件下,週期是一個定值
所以先燒其中一條繩子,用另一條繩子掛重物讓它擺
測量當第一條燒完的時候(一個小時),第二條會擺幾下
舉個例子,假設一小時是擺3600下,那週期大概就是一秒鐘
我想要量45分鐘,就讓它擺60*45下
--
鐘擺週期公式是:T=2*pi*sqrt(L/g)
重點是,在擺幅不大、繩子長短固定的條件下,週期是一個定值
所以先燒其中一條繩子,用另一條繩子掛重物讓它擺
測量當第一條燒完的時候(一個小時),第二條會擺幾下
舉個例子,假設一小時是擺3600下,那週期大概就是一秒鐘
我想要量45分鐘,就讓它擺60*45下
--
Tags:
推理遊戲
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