物理振动波动学解答题
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速度分布：v(x,t)=-Aωsin[ω(t-x/u)+Φ]
由题意：t=0 x=0时： y=0 v>0
则 Φ=-π/2
所以 x=0处质点振动方程：y=Acos(ωt-π/2)
(2)波动方程（波函数）：y(x,t)=Acos[ω(t-x/u)-π/2]
其中 A=0.1m u=1m/s ω=π
按部就班地计算
绛旓細锛1锛娉㈠姩鏂圭▼锛歽(x,t)=Acos[蠅(t-x/u)+桅]閫熷害鍒嗗竷锛歷(x,t)=-A蠅sin[蠅(t-x/u)+桅]鐢遍鎰忥細t=0 x=0鏃讹細 y=0 v>0 鍒 桅=-蟺/2 鎵浠 x=0澶勮川鐐鎸姩鏂圭▼锛歽=Acos(蠅t-蟺/2)(2)娉㈠姩鏂圭▼锛堟尝鍑芥暟锛夛細y(x,t)=Acos[蠅(t-x/u)-蟺/2]鍏朵腑 A=0.1m u=...
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