tabogirl910 2010-7-30 11:21 AM
phy mc 97 #8
答案岩,但我的做法同MARKING唔同
可以睇下我有冇錯嗎?
同埋我唔明MARKING的做法......
thxem46
[[i] 本帖最後由 tabogirl910 於 2010-7-30 12:45 PM 編輯 [/i]]
tabogirl910 2010-7-30 12:46 PM
回覆 #2 傑Simon 的帖子
唔好意思急得滞忘記左POST....
Simon 2010-7-30 01:51 PM
應該都岩既
不過最好標明係v[subtag]max[/subtag] = ωA
因為max. K.E. 搵到既係max. v
tabogirl910 2010-7-30 07:40 PM
但我唔明MARKING的3x10^-3=1/2kA^2
Simon 2010-7-30 08:23 PM
Max. K.E. = Max. P.E.
因為max. K.E. => P.E. = 0
max. P.E. => K.E. = 0
有錯請指吧!!!
tabogirl910 2010-7-30 09:18 PM
回覆 #6 傑Simon 的帖子
icic....thx
另外,我的做法咪代左X=0....
呢題係好耐之前做...我忘了點解我會代X=0...
kiwakwok 2010-7-31 09:58 AM
Hope that it helps
[attach]28547[/attach]
In the picture, the left part is what you have seen from the marking scheme, while the right part is what you have done.
The formulae in light blue are inter-convertible, and the formula in red is just the conversion factor.
Therefore, both methods are okay as they are more or less the same.
Note: Why substitute x = 0? For maximum velocity, the mass is at the equilibrium position.
tabogirl910 2010-7-31 10:48 AM
回覆 #8 kiwakwok 的帖子
it is of great use to me !
thanks a lot !