When two progressive waves y1-4sin2x - 6t and y2-3sin2x-6t-2 are superimposed- the amplitude of the resultant wave is

Resultant amplitude = √(A_1^2 + A_2^2 + 2 A_1A_2cosθ) phase difference = θ = π2 #160; = √(4^2 + 3^2) =5 ...

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YDSE Problems II

When two prog...

Question

When two progressive waves $y_{1} = 4 sin (2 x - 6 t)$ and $y_{2} = 3 sin (2 x - 6 t - \frac{π}{2})$ are superimposed, the amplitude of the resultant wave is

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y_{1} = A_{1} sin (ω t - β_{1})

y_{2} = A_{2} sin (ω t - β_{2})

Y_{1}

Y_{2}

(c o t + ϕ)

ϕ = \frac{π}{2}

Two waves $y_{1} = A_{1} sin (ω t - β_{1}) y_{2} = A_{2} sin (ω t - β_{2})$ superimpose to form a resultant wave whose amplitude is

y_{1} = A sin (k x - ω t)

y_{2} = A cos (k x - ω t)

y_{1} = A sin (ω t - k x)

y_{2} = A sin (ω t - k x + ϕ)

\sqrt{2} A

ϕ

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