作者kramnik (progressive)
看板Physics
标题Re: [题目] 简谐运动
时间Tue Jan 12 12:33:51 2010
※ 引述《nerv3890 (阿渊)》之铭言:
: [领域] (题目相关领域)
: 简谐
: [来源] (课本习题、考古题、参考书...)
: 考古
: [题目]
: 3.A pendulum of length L and mass M has a spring of force constant k
: connected to it at a distance h below its point of suspension as shown in a
: figure below. Find the period of vibration of the system if the amplitude of
: vibration is small. Assume the vertical suspension of length L is rigid and
: its mass is negligible.
: _____________________
: | |
: |h |
: | |
: L |__ /\ ____|
: | \/ \/ |
: | k |
: O
: M
: [瓶颈] (写写自己的想法,方便大家为你解答)
: 初步构想
: 是设力矩τ=-CΘ
: Θ为摆动角度,C为常数
: 之後就都没想法了...Orz
: 有人能帮解答吗
: 感恩
如下图所示..以悬挂点作为支点..设系统角加速度为α..顺时针为正..
http://www.wretch.cc/album/show.php?i=kramnik1&b=34&f=1825180178&p=11
因为"Assume the vertical suspension of length L is rigid"..
根据力矩守恒
M*g*sinθ*L + F(spring)*cosθ*h = -M*L^2*α ..............(a)
因为"the force constant of spring is k"
F(spring) =k*h*sinθ .....................................(b)
由(a)(b)知
M*g*sinθ*L + k*h^2*sinθ*cosθ = -M*L^2*α ..............(c)
因为"the amplitude of vibration is small"
对(c)做θ泰勒展开1阶近似..可得
M*g*L*θ + k*h^2*θ = -M*L^2*α
θ = -M*L^2/(M*g*L + k*h^2)*α.............................(d)
(d)式之解为
θ = A^(i*w*t) ,其中 w = M*L^2/(M*g*L + k*h^2)
设The period of vibration of the system为T..则
T = 2*π/w
= 2*π*(M*g*L + k*h^2)/(m*L^2)
--
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◆ From: 118.168.83.172
1F:推 nerv3890:k大 那个倒数第4行 那个A是什麽 Q___Q 01/12 23:05
2F:推 colorya2001:α与θ的关系式不是 α=-w^2θ吗? 这样求出的w与原po 01/13 08:09
3F:→ colorya2001:的w就会不一样 01/13 08:10
4F:→ kramnik:A是比例常数..需要起始条件才可以解.. 01/13 10:27