Answer to Question #195533 in Mechanics | Relativity for Mohammad Balhas

Question

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

Answer 1

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

$\phi_o$ : Phase difference

Resultant wave, $y=y_1+y_2$

$y=A\sin(kx-\omega t)+A\sin(kx-\omega t+\phi_o)$

$y=A\sin(kx\pm\omega t+\alpha)$

$\tan \alpha=\dfrac{\sin\phi_o}{1+\cos\phi_o}$

From question,

$A=6\space cm$

$\lambda=\dfrac{2}{3}\space m$

$v=50\space m/s$

$\omega=\dfrac{2\pi v}{\lambda}=150\pi$

$k=\dfrac{2\pi}{\lambda}=3\pi$

$\phi_o=$ Phase constant = $3\pi$

$\tan \alpha=\dfrac{\sin3\pi}{1+\cos3\pi}$

$\alpha=\tan^{-1}\infin$

$\alpha=\dfrac{\pi}{2}$

As wave travels in left direction equation of wave will be

$y=A\sin(kx+\omega t+\alpha)$

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