ξ=Asin(ωt−kx+ϕ0)\xi=A\sin(\omega t-kx+\phi_0)ξ=Asin(ωt−kx+ϕ0)
y=25sin(120t−40x)y=25\sin(120 t-40x)y=25sin(120t−40x)
We have
a) k=2π/λ→λ=2π/k=2⋅3.14/40=0.157(m)k=2\pi/\lambda\to \lambda=2\pi/k=2\cdot3.14/40=0.157(m)k=2π/λ→λ=2π/k=2⋅3.14/40=0.157(m)
b) k=ω/v→v=ω/k=120/40=3(m/s)k=\omega/v\to v=\omega/k=120/40=3(m/s)k=ω/v→v=ω/k=120/40=3(m/s) (phase velocity)
c) ω=2πν→ν=ω/(2π)=120/(2⋅3.14)=19.1(Hz)\omega=2\pi\nu\to\nu=\omega/(2\pi)=120/(2\cdot3.14)=19.1(Hz)ω=2πν→ν=ω/(2π)=120/(2⋅3.14)=19.1(Hz)
d) T=1/ν=1/19.1=0.052(s)T=1/\nu=1/19.1=0.052(s)T=1/ν=1/19.1=0.052(s)
e) A=25(m)A=25(m)A=25(m)
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments