The phase difference between two waves, represented by y 1 = 1 0 − 6 sin { 100 t + ( x /50 ) + 0.5 } m y 2 = 1 0 − 6 cos { 100 t + ( 50 x ) } m where x is expressed in meters and t is expressed in seconds, is approximately:-

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The phase difference between two waves, represented by $y_{1} = 1 0^{- 6} sin {100 t + (x /50) + 0.5} m$

$y_{2} = 1 0^{- 6} cos {100 t + (\frac{x}{50})} m$

where x is expressed in meters and t is expressed in seconds, is approximately:-

Answer:

Solution:

To find the phase difference, first convert $y_{2}$ to a sine function: $cos (θ) = sin (θ + π /2)$ .

So, $y_{2} = 1 0^{- 6} sin {100 t + (x /50) + π /2} m$ .

The phase of $y_{1}$ is $ϕ_{1} = 100 t + (x /50) + 0.5$ .

The phase of $y_{2}$ is $ϕ_{2} = 100 t + (x /50) + π /2$ .

The phase difference is $Δ ϕ = ∣ ϕ_{2} - ϕ_{1} ∣ = ∣ π /2 - 0.5∣ = ∣1.57 - 0.5∣ = 1.07$ rad.

Stream:JEESubject:PhysicsTopic:Wave on StringSubtopic:Equation/graph

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