Answer:
The phase difference between yA and yB is [tex]\frac{-\pi }{12}[/tex]
Step-by-step explanation:
Given harmonic modeled as :
yA = 8 sin(2t - [tex]\frac{\pi }{3}[/tex]) And
yB = 8 sin(2t - [tex]\frac{\pi }{4}[/tex])
The function as written as :
y = a sin(ωt - Ф) where Ф is phase difference
So , phase difference between yA and yB = ( Ф_1 - Ф_2 )
Or phase difference between yA and yB = ( - [tex]\frac{\pi }{3}[/tex] + [tex]\frac{\pi }{4}[/tex] )
Or, phase difference between yA and yB = [tex](\frac{-4\pi +3\pi }{12})[/tex]
I.e phase difference between yA and yB =[tex]\frac{-\pi }{12}[/tex]
Hence The phase difference between yA and yB is [tex]\frac{-\pi }{12}[/tex] Answer
Answer:
-π/12
Step-by-step explanation:
yA has a phase shift of -π/3; yB has a phase shift of -π/4. The difference between their phases is ...
-π/3 -(-π/4) = (-4/12 +3/12)π = -π/12
yA lags yB by π/12 radians.
