Respuesta :
Answer:
Therefore the amplitude of the resultant wave is [tex]=0.95 y_m[/tex]
Explanation:
The equation of wave:
y=A sin (kx-ωt)
For wave 1:
y₁=A sin (kx-ωt) = [tex]y_{m}[/tex]sin (kx-ωt)
For wave 2:
y₂=A sin (kx-ωt+Φ) = [tex]y_{m}[/tex]sin (kx-ωt+Φ)
Where A= amplitude=[tex]y_m[/tex]
The angular frequency [tex]\omega=\frac{2\pi}{T}[/tex]
[tex]k=\frac{2\pi}{\lambda}[/tex] , [tex]\lambda[/tex]= wave length.
t= time
T= Time period
[tex]\phi[/tex] = phase difference = [tex]\frac{\pi}{5}[/tex]
The resultant wave will be
y = y₁ + y₂
=[tex]y_m[/tex] sin (kx-ωt) + [tex]y_m[/tex] sin (kx-ωt+Φ)
[tex]=y_m[/tex] {sin (kx-ωt) + sin (kx-ωt+Φ)}
[tex]=y_m\ sin(\frac{kx-\omega t +\phi + kx-\omega t }2)\ cos(\frac{kx-\omega t +\phi -kx+\omega t}2)[/tex]
[tex]=y_m\ sin({kx-\omega t +\frac\phi 2)\ cos(\frac{\phi }2)[/tex]
[tex]=y_m\ cos(\frac{\phi }2) sin({kx-\omega t +\frac\phi 2)[/tex]
Therefore the amplitude of the resultant wave is
[tex]=y_m\ cos(\frac{\phi }2)[/tex]
[tex]=y_m\ cos(\frac{\pi }{10})[/tex]
[tex]=0.95 y_m[/tex]
Answer:
- Total amplitude/common amplitude = [tex]1.902[/tex]
Explanation:
[tex]y1 = ym*sin(kx-wt)\\\\y2 = ym*sin(kx-wt-\pi/5)[/tex]
[tex]ytotal = y1+y2\\\\ytotal = ym*sin(kx-wt) + ym*sin(kx-wt-\pi/5)\\\\sin(A+B) = 2*sin(A+B /2)*cos(A-B/2)\\\\here A = kx-wt\\\\\And B = kx-wt-\pi/5\\\\[/tex]
[tex]ytotal = 2*ym*sin((1/2)*(2kx-2wt-\pi/5))*cos(\pi/10)\\\\ytotal = 2*0.951*ym*sin((1/2)*(2kx-2wt-\pi/5))\\\\ytotal = 1.902*ym*sin(kx-wt-\pi/10))[/tex]
amplitude of the resultant wave is ytotal is 1.902 times common amplitude
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