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Doppler Effect in Light Waves

Two waves are...

Question

Two waves are propagating along a taut string that coincides with the x-axis. The first wave has the wave function $y_{1} = A c o s [k (x - v t)]$ and the second has the wave function $y_{2} = A c o s [k (x + v t) + ϕ] .$

For constructive interference at

x = 0, ϕ = π .

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For constructive interference at

x = 0, ϕ = 3 π .

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For destructive interference at

x = 0, ϕ = π .

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For destructive interference at

x = 0, ϕ = 2 π

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Solution

The correct option is C For destructive interference at $x = 0, ϕ = π .$
Resultant wave function $y = y_{1} + y_{2}$

$y = A c o s [k (x - v t)] + A c o s [k (x + v t) + ϕ]$

Using $c o s A + c o s B = 2 c o s \frac{A + B}{2} c o s \frac{A - B}{2}$

$y = 2 A c o s [\frac{k (x - v t) + k (x + v t) + ϕ}{2}] c o s [\frac{k (x - v t) - k (x + v t) + ϕ}{2}]$

$y = 2 A c o s [k x + \frac{ϕ}{2}] c o s [\frac{ϕ}{2} - v t]$

Now for destructive interference , $y = 0$ i.e $c o s [k x + \frac{ϕ}{2}]$ must be equal to zero.

For $x = 0, ϕ = π ⟹ c o s [k x + \frac{ϕ}{2}] = 0$

For constructive interference , $y$ must be maximum i.e $c o s [k x + \frac{ϕ}{2}] = 1$ OR $- 1$ .

For $x = 0, ϕ = 2 π ⟹ c o s [k x + \frac{ϕ}{2}] = - 1$

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Two waves are propagating along a taut string that coincides with the x-axis. The first wave has the wave function y1=Acos[k(x−vt)] and the second has the wave function y2=Acos[k(x+vt)+ϕ].

Two waves are propagating along a taut string that coincides with the x-axis. The first wave has the wave function $y_{1} = A c o s [k (x - v t)]$ and the second has the wave function $y_{2} = A c o s [k (x + v t) + ϕ] .$