Answer:
two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is yR (x, t) = 0.70 m sin⎛ ⎝3.00 m−1 x − 6.28 s−1 t + π/16 rad⎞ ⎠ . What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?
ω = 6.28 s − 1 ,
k = 3.00 m− 1 ,
φ = π rad,
A R = 2 A cos (φ 2 ) ,
A = 0.37 m
Step-by-step explanation:
y1 ( x , t ) = A sin( k x − ω t +φ ) ,
y 2 ( x , t ) = A sin ( k x − ω t ) .
from the principle of superposition which states that when two or more waves combine, there resultant wave is the algebriac sum of the individual waves
y1 ( x , t ) = A sin( k x − ω t +φ ) , is generaL form of thw wave eqaution
A=amplitude
k=angular wave number
ω=angular frequency
φ =phase constant
k=2π/lambda
ω=2π/T
yR (x, t) = 0.70 m sin{3.00 m−1 x − 6.28 s−1 t + π/16 rad}....................*
two waves superposed to give the above, assuming they are moving in the +x direction
y1 ( x , t ) = A sin( k x − ω t +φ ) , .....................1
y 2 ( x , t ) = A sin ( k x − ω t ) ...........................2
adding the two equation will give
A sin( k x − ω t +φ )+A sin ( k x − ω t ) .................3
A( sin( k x − ω t +φ )+ sin ( k x − ω t ) ),......................4
similar to the following trigonometry identity
sina+sinb=2cos(a-b)/2sin(a+b)/2
let a= ( k x − ω t
b=k x − ω t +φ )
y(x,t)=2Acos(φ/2)sin(k x − ω t +φ/2)
k=3m^-1
lambda=2π/k=2.09m
ω=6.28= T=2π/6.28
T=1s
φ/2=π/16
φ=π/8rad
amplitude
2Acos(φ/2)=0.70 m
A=0.7/2cos(π/8)
A=0.37 m