Answer to Question #206294 in Mechanics | Relativity for KABIR

Question

• 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?

Answer 1

(1) The wave function of this wave?

$y=Asin(kx± \omega t + \phi)$

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

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

$1=sin(\phi)$

$\phi = \frac{\pi}{2}$

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

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

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

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

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

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

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

$y=Asin( \frac{\pi}{2})$

$y=A=0.1 m$

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

$\frac{d}{dx}\left(A\sin \left(\frac{\pi }{2}\right)\right)$

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

$v=0m/s$