See below and explain fully. The superposition y(x,t) = y1(x,t) + y2 (x,t) of tw

Question

See below and explain fully.

The superposition y(x,t) = y1(x,t) + y2 (x,t) of two harmonic waves y1 = A sin(kx-omegat-Phi1) y2 = A sin(kx-omegat-Phi2) where A = 5.6cm, k: = 2.3m-1, omega=6.1s-1, Phi1=4.53 rad, and Phi2=-0.911 rad is itself a harmonic wave y=A' sin(k'x-omega't-Phi'). Consider the time-plot of the resultant wave y(t) at fixed position x = 0: What is the angular frequency of this resultant wave? A useful trig identity is sin alpha+ sin beta=2 sin alpha+beta/2 cos alpha-beta/2 Answer in units of s-1 Consider the supersition wave y(x) at time t = 0: What is the wavelength of this resultant wave? Answer in units of m What is the initial phase of the resultant wave (as an angle in radians between 0 and 2 7pi)? Answer in units of rad What is the amplitude of the resultant wave? Answer in units of cm What is the displacement of the resultant wave at t = 6 s and x = 8m? Answer in units of cm

Explanation / Answer

sin a + sin b = 2 sin (a + b)/2 cos(a-b)/2

a + b = 2 kx - 2 wt - theta 1 - theta 2

a-b = theta2 - theta 1

so we have

y = A sin( kx - wt - (theta 1 + theta2)/2 ) cos( (theta2-theta1)/2)

so w' = w=6.1 rad/s

7) wavelength = 2 pi/k = 2*pi/2.3= 2.73 m

8) phase = (theta1 + theta2)/2 = (4.52 - 0.911)/2=1.805

9) A' = A cos((theta 2 - theta1)/2) = 5.6*cos( (4.52 + 0.911)/2) = 5.1

10)

y = 5.6*cos( (4.52 + 0.911)/2)*sin( 2.3*8 - 6.1*6 - 1.805)=4.67 cm

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