Question #206294
  • A sinsoidal wave with amplitude A= 0.1m, period T=1/2 s, and wave length λ=10m. When t=0, the element at the wave source happened to be at the positive maximum displacement position. If the wave source is the origin point of the x axis and the wave is travelling along the positive OX axis, find: (1) The wave function of this wave? (2) The displacement of the element at x₁=λ/4 when T₁=T/4? (3) The velocity of the element at x₁=λ/4 when T₂=T/2?
1
Expert's answer
2021-06-16T14:26:59-0400

(1) The wave function of this wave?

y=Asin(kx±ωt+ϕ)y=Asin(kx± \omega t + \phi)

Given x=0 and t= 0, y = A

y=Asin(k0ω0+ϕ)y=Asin(k*0- \omega *0+ \phi)

1=sin(ϕ)1=sin(\phi)

ϕ=π2\phi = \frac{\pi}{2}

Therefore the equation becomes y=Asin(kxωt+π2)y=Asin(kx-\omega t + \frac{\pi}{2})

y=Asin(kxωt+π2)y=Asin(kx- \omega t + \frac{\pi}{2})

y=Asin(2πλx2πTt+π2)y=Asin( \frac{2 \pi}{\lambda}x- \frac{2 \pi}{T} t + \frac{\pi}{2})

y=Asin(π5x4πt+π2)y=Asin( \frac{ \pi}{5}x- 4 \pi t + \frac{\pi}{2})

(2) The displacement of the element at x₁=λ/4 when T₁=T/4?

y=Asin(2πλλ42πTT4+π2)y=Asin( \frac{2 \pi}{\lambda}*\frac{ \lambda}{4}- \frac{2 \pi}{T}* \frac{T}{4} + \frac{\pi}{2})

y=Asin(π2π2+π2)y=Asin( \frac{\pi}{2}- \frac{\pi}{2} + \frac{\pi}{2})

y=Asin(π2)y=Asin( \frac{\pi}{2})

y=A=0.1my=A=0.1 m

(3) The velocity of the element at x₁=λ/4 when T₂=T/2?

ddx(Asin(π2))\frac{d}{dx}\left(A\sin \left(\frac{\pi }{2}\right)\right)

ddx(A)=0\frac{d}{dx}\left(A\right) =0

v=0m/sv=0m/s


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