Question #195533

wo sinusoidal waves of wavelength λ = 2/3 m and amplitude A = 6 cm and differing with their phase constant, are travelling to the right with same velocity v = 50 m/s. The resultant wave function y_res (x,t) will have the form:


y_res (x,t) = 12(cm) cos⁡(φ/2) sin⁡(150πx-3πt+φ/2).


y_res (x,t) = 12(cm) cos⁡(φ/2) sin⁡(3πx+150πt+φ/2).


y_res (x,t) = 12(cm) cos⁡(φ/2) sin⁡(150πx+3πt+φ/2).


y_res (x,t) = 12(cm) cos⁡(φ/2) sin⁡(3πx-180πt+φ/2).


y_res (x,t) = 12(cm) cos⁡(φ/2) sin⁡(3πx-150πt+φ/2).

Clear selection

Two identical sinusoidal waves with w


1
Expert's answer
2021-05-24T16:48:15-0400

Let two waves, y1=Asin(kxωt)y_1=A\sin(kx-\omega t) and y2=Asin(kxωt+ϕo)y_2=A\sin(kx-\omega t+\phi_o) be travelling in the same direction,

ϕo\phi_o : Phase difference

Resultant wave, y=y1+y2y=y_1+y_2

y=Asin(kxωt)+Asin(kxωt+ϕo)y=A\sin(kx-\omega t)+A\sin(kx-\omega t+\phi_o)

y=Asin(kx±ωt+α)y=A\sin(kx\pm\omega t+\alpha)

tanα=sinϕo1+cosϕo\tan \alpha=\dfrac{\sin\phi_o}{1+\cos\phi_o}

From question,

A=6 cmA=6\space cm

λ=23 m\lambda=\dfrac{2}{3}\space m

v=50 m/sv=50\space m/s

ω=2πvλ=150π\omega=\dfrac{2\pi v}{\lambda}=150\pi

k=2πλ=3πk=\dfrac{2\pi}{\lambda}=3\pi

ϕo=\phi_o= Phase constant = 3π3\pi

tanα=sin3π1+cos3π\tan \alpha=\dfrac{\sin3\pi}{1+\cos3\pi}

α=tan1\alpha=\tan^{-1}\infin

α=π2\alpha=\dfrac{\pi}{2}

As wave travels in left direction equation of wave will be

y=Asin(kx+ωt+α)y=A\sin(kx+\omega t+\alpha)

y=6sin(3πx+150πt+π2)y=6\sin\bigg(3\pi x+150\pi t+\dfrac{\pi}{2}\bigg)


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