Question

Y1=a1 sin (wt-kx) and y2=a2cos (wt-kx+#) path difference is

Answer 1

Answer:

y1 = a sin(wt-kx) and y2 = acos(wt-kx)

y2 = a sin(wt-kx+π/2)

so, according to the superposition principle,

y = y1+y2

y = a sin(wt-kx)+a sin(wt-kx+π/2)

use Trignometric formula sin a + sin b = 1/2sin((a+b)/2)cos((a-b)/2)

y = a*(2sin*1/2(wt-kx+wt-kx+π/2)cos*1/2(wt-kx-wt+kx-π/2))

y = a *2*sin1/2(2*(wt-kx+π/4)cos(-π/4)

y = a*2*sin(wt-kx+π/4)*1/√2

y = √2a sin(wt-kx+π/4)

so amplitude is √2a and phase difference is π/4.

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Two waves whether superposed or not, is not needed. The phase difference between two waves….

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You dont need to superpose waves to find the phase difference. Here,

y1=asin(ωt−kx); y2=acos(ωt−kx)

⟹y1=asin(ωt−kx);

y2=asin(π2−(ωt−kx))

So difference in phase is π2