一平面简谐波的表达式为y=0.025cos(125t–0.37x),求其角频率w,波速v和波长a,

2025-05-18 18:03:21
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回答1:

y = 0.025 cos(125t – 0.37x)
y = 0.025 cos(ω t – ω x / v)
y = 0.025 cos(ω t – 2π x / λ)
角频率 ω = 125 /s
ω/v = 0.37,波速 v = ω/0.37 = 337.8 m/s
2π / λ = 0.37,波长 λ = 2π / 0.37 = 17.0 m