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單自由度運動系統 PowerPoint PPT Presentation


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河工 4A B94520021  余東軒. 單自由度運動系統. 已知 F=Ma=Mx’’(t)=Kx(t) 單自由度,設 v ={x(t)}={x}cos ω t or v ={x}sin ω t , 且 v ≠0 ,以 v 代入原式之 x(t) 。 K v = ω 2 M v , ω 2 M v -K v =0 , det[ ω 2 M-K]=0 其中 K 為勁度矩陣, M 為質量矩陣。. 由上圖,物體將承受左彈簧的張力和右彈簧之拉力,即 (k+k) v =2k v =ω 2 m v 。

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單自由度運動系統

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3700319

河工4AB94520021 余東軒

單自由度運動系統

已知F=Ma=Mx’’(t)=Kx(t)

單自由度,設v={x(t)}={x}cosωtorv={x}sinωt,

且v≠0,以v代入原式之x(t)。

Kv=ω2Mv,ω2Mv-Kv=0,det[ω2M-K]=0

其中K為勁度矩陣,M為質量矩陣。


3700319

  • 由上圖,物體將承受左彈簧的張力和右彈簧之拉力,即(k+k)v=2kv=ω2mv。

  • ω2m-2k=0,ω2m=2k,ω=(2k/m)1/2=1.41421(k/m)1/2

  • x(t)=cos{(2)1/2*t}=cos(1.41421*t)。

    由於在座標軸上,向左位移時x為負值,因此在Mathematica上運算時:

    需將左彈簧的右端和右彈簧的左端x(t)設為-cos(1.41421*t),固定邊不需移動,則設為0。


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