一平面简谐波的波动方程为y=0.02cosπ(50t+4x)(SI),此波的振幅

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问题 一平面简谐波的波动方程为y=0.02cosπ(50t+4x)(SI),此波的振幅和周期分别为(  )。A.0.02cm,0.04sB.0.02m,0.02sC.-0.02m,0.02sD.0.02m,25s

选项 A.0.02cm,0.04s
B.0.02m,0.02s
C.-0.02m,0.02s
D.0.02m,25s

答案 A

解析 平面简谐波的波动方程为:y=Acos[ω(t-x/u)+φ0],对比可知,振幅A=0.02m,ω=50π。由T=2π/ω可得,T=0.04s。
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