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單自由度運動系統

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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  1. 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。

  2. 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。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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