Question

设有一沿x轴正向传播的波,其波动方程为。写出波源的振动方程。

Answer 1

问：设有一沿x轴正向传播的波,其波动方程为。写出波源的振动方程。

答：由波源的振动方程为y=0.3cos200πt 得到w=200π周期T=2π/w=1/100（S）其波长为3m,求得波速ν=3/T=300m/s波动方程设为y=0.3cosw(t-Δt)波从原点传播X的距离,原点的波传到此点所用时间Δt=x/ν=x/300波动方程设为y=0.3cosw(t-Δt)=0.3cosw(t-x/300) =0.3cos200π(t-x/300)

问：某人手拿铅球，肘关节为支点受力图如下：其中前臂重W为20N，臂长（包括手）L为0.5m，L1为0.05m，L2为0.2m。铅球重R为10N。角α为45°。求手臂静止时，肌肉用力F

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答：你学习这个模型，你应该就会做这类型的题目。