Question

Two shm's are respectively represented by y=a sin((omega)*t-kx) and y=b cos((omega)*t-kx). find out the phase difference between the two?

Answer 1

y = a sin (wt-kx)   & 
y = b cos (wt-kx)  or b sin (wt -kx + 90o)
so the phase difference is 90o

Answer 2

phase difference is π/2

Two SHMs are respectively represented by

y = a sin(ωt - kx) and y = b cos(ωt - kx)

we have to find phase difference,

first of all, you have to convert both of them in the trigonometrical equation.

I mean, either you should convert sin to cos of the first or you can convert convert cos to sin of the 2nd wave equation.

let's change 2nd one !

y = b cos(ωt - kx) = b sin(ωt - kx + π/2) [ we know, sin(π/2 + x) = cosx ]

now phase difference, ∆Φ = (ωt - kx + π/2) - (ωt - kx) = π/2

also read similar questions: determine whether sin square omega t represent SHM,periodic but not SH.give the period.

https://brainly.in/question/1178712

the displacement wave is represented by y=Asin(wt-kx) x is the distance t is the time

write dimensional formula for w a...

https://brainly.in/question/1101413