Question

Two identical traveling waves, moving in the same direction, are out of phase by π/5.0 rad. What is the amplitude of the resultant wave in terms of the common amplitude ym of the two combining waves? (Give the answer as the ratio of the total amplitude to the common amplitude.)

Answer 1

Answer:

Therefore the amplitude of the resultant wave is $=0.95 y_m$

Explanation:

The equation of wave:

y=A sin (kx-ωt)

For wave 1:

y₁=A sin (kx-ωt) = $y_{m}$sin (kx-ωt)

For wave 2:

y₂=A sin (kx-ωt+Φ) = $y_{m}$sin (kx-ωt+Φ)

Where A= amplitude=$y_m$

The angular frequency $\omega=\frac{2\pi}{T}$

$k=\frac{2\pi}{\lambda}$ , $\lambda$= wave length.

t= time

T= Time period

$\phi$ = phase difference =  $\frac{\pi}{5}$

The resultant wave will be

y = y₁ + y₂

 =$y_m$ sin (kx-ωt) + $y_m$ sin (kx-ωt+Φ)

 $=y_m$ {sin (kx-ωt) + sin (kx-ωt+Φ)}

 $=y_m\ sin(\frac{kx-\omega t +\phi + kx-\omega t }2)\ cos(\frac{kx-\omega t +\phi -kx+\omega t}2)$

 $=y_m\ sin({kx-\omega t +\frac\phi 2)\ cos(\frac{\phi }2)$

$=y_m\ cos(\frac{\phi }2) sin({kx-\omega t +\frac\phi 2)$

Therefore the amplitude of the resultant wave is

$=y_m\ cos(\frac{\phi }2)$

$=y_m\ cos(\frac{\pi }{10})$

$=0.95 y_m$