大学物理题目 大学物理题
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2\u3001\u5e76\u8054\u7535\u8def
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P=W/t
dv/dt=-kv³
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波动方程：y=Acos[w(t+x/u)+φ] 式中 w=2πf=500π
由题意可知，当x=0 t=0 时，y=√2A/2 所以φ=π/4
则：y=Acos[w(t+x/u)+π/4] 你这里没给出波长，所以求不出 波速u
有波速 u 只要将 x=100 代入 上式就可以得到 距离原点100m处质点的振动方程。振动方程对时间t求一阶导数 就得到 速度了
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