大学物理波动问题
\u5927\u5b66\u7269\u7406\u5173\u4e8e\u6ce2\u52a8\u7684\u9898\u76eeO\u70b9\u7684\u632f\u52a8\u65b9\u7a0b\u4e3a yO = Acos(2\u03c0\u03bdt)\uff1b
\u5165\u5c04\u6ce2\u5728\u8ddd\u6ce2\u6e90d\u5904\u7684\u632f\u52a8\u65b9\u7a0b\u4e3a y\u5165d = Acos(2\u03c0\u03bdt - 2\u03c0d/\u03bb ) \uff1b
\u53cd\u5c04\u6ce2\u5728\u8ddd\u6ce2\u6e90d\u5904\u7684\u632f\u52a8\u65b9\u7a0b\u4e3a y\u53cdd = Acos(2\u03c0\u03bdt - 2\u03c0d/\u03bb ) ,\u5373\u56e0\u4e3a\u6ca1\u6709\u534a\u6ce2\u635f\u5931\u4e0e\u4e0a\u5f0f\u76f8\u540c\uff1b
\u53cd\u5c04\u6ce2\u7684\u6ce2\u52a8\u65b9\u7a0b\u4e3a y\u53cd = Acos(2\u03c0\u03bdt - 2\u03c0d/\u03bb - 2\u03c0(d-x)/\u03bb )
\u6240\u4ee5\uff0c\u9009 C
\u628a\u6bcf\u7ec4\u7684y1\u4e0ey2\u76f8\u52a0\uff0c\u5982\u679c\u548c\u4e3a\u9898\u76ee\u4e2d\u7684\u9a7b\u6ce2\u8868\u8fbe\u5f0f\uff0c\u90a3\u8fd9\u4e00\u7ec4\u5c31\u662f\u7b26\u5408\u6761\u4ef6\u7684\u3002
\u4ee5\u7b2c\u4e00\u7ec4\u4e3a\u4f8b\uff1a
y1=0.1cos[2\u03c0(10t-x)+\u03c0/2]=-0.1sin[2\u03c0(10t-x)]=-0.1sin20\u03c0tcosx+cos20\u03c0tsinx
\u540c\u7406\uff1a
y2=0.1cos[2\u03c0(10t+x)+\u03c0/2]=-0.1sin[2\u03c0(10t+x)]=-0.1sin20\u03c0tcosx-cos20\u03c0tsinx
\u5219y1+y2=-0.2sin20\u03c0tcosx \u4e0e\u5df2\u77e5\u6761\u4ef6\u4e0d\u7b26\uff0c\u6240\u4ee5A\u4e0d\u662f\u3002
\u4f9d\u6b64\u7c7b\u63a8\uff0c\u53ef\u4ee5\u8ba1\u7b97\u51faC\u662f\u7b26\u5408\u6761\u4ef6\u7684\uff0c\u6240\u4ee5\u7b54\u6848\u662fC\u3002
故有：y=Acos（2πvt+φ），
当：t=t‘，且质点运动放心向下，则有：0=Acos（2πvt‘+φ），且：2πvt‘+φ=π/2，
解得：φ=π/2-2πvt‘
则有质点振动方程为：y=Acos（2πvt+π/2-2πvt‘）=Acos[2πv(t-t')+π/2]
2、已知频率v，波速u
则有：谐波的基本表达式为：y=Acos[2πv(t-x/u)+π/2-2πvt‘]=Acos[2πv(t-t'-x/u)+π/2]，
ω=2πν,t=t',y=0且向负方向运动，φ0=π/2
振动方程
y(t)=Acos[ω(t-t')+π/2]=Acos[2πν(t-t')+π/2]
波动方程
y(t)=Acos[ω(t-t'-x/u)+π/2]=Acos[2πν(t-t'-x/u)+π/2]
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