If three waves- y1-Asinkx- t-y2-Asinkx- t- and y3-Asin kx- t-2- are superimposed so that y1-y2-y3-0 at al positions- Then the value of - will be

Y=y_1+y_2+y_3=A[sin(kx ω t)+sin(kx ω t+ϕ)+sin (kx ω t+2ϕ)]=0 ⇒sin(kx ω t)+sin(kx ω t+ϕ)+sin(kx ω t+2ϕ)=0 nbsp; ⇒ 2sin(kx ω t)sinϕ/2+2sin(kx ω t+ϕ)sinϕ/2+2s ...

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Byju's Answer

Standard XII

Physics

Phasor Diagrams in AC Circuits

If three wave...

Question

If three waves, $y_{1} = A sin (k x - ω t), y_{2} = A sin (k x - ω t + ϕ)$ and $y_{3} = A sin (k x - ω t + 2 ϕ)$ are superimposed so that $y_{1} + y_{2} + y_{3} = 0$ at al positions. Then the value of $ϕ$ will be

- \frac{2 π}{3}

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All of these

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\frac{4 π}{3}

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\frac{2 π}{3}

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Solution

The correct option is B All of these
$y = y_{1} + y_{2} + y_{3} = A [sin (k x - ω t) + sin (k x - ω t + ϕ) + sin (k x - ω t + 2 ϕ)] = 0$

$\Rightarrow sin (k x - ω t) + sin (k x - ω t + ϕ) + sin (k x - ω t + 2 ϕ) = 0$

$\Rightarrow 2 sin (k x - ω t) sin \frac{ϕ}{2} + 2 sin (k x - ω t + ϕ) sin \frac{ϕ}{2} + 2 sin (k x - ω t + 2 ϕ) = 0$

$\Rightarrow cos (k x - ω t - \frac{ϕ}{2}) - cos (k x - ω t + \frac{3 ϕ}{2}) + cos (k x - ω t - \frac{3 ϕ}{2}) - cos (k x - ω t - \frac{5 ϕ}{2})$
$\Rightarrow cos (k x - ω t - \frac{ϕ}{2}) = cos (k x - ω t + \frac{5 ϕ}{2})$

$\Rightarrow \frac{5 ϕ}{2} = 2 n π \pm \frac{ϕ}{2}$

$∴ ϕ = \frac{2 π}{3} or \frac{- 2 π}{3} or \frac{4 π}{3}$

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If three waves, y1=Asin(kx−ωt),y2=Asin(kx−ωt+ϕ) and y3=Asin(kx−ωt+2ϕ) are superimposed so that y1+y2+y3=0 at al positions. Then the value of ϕ will be

If three waves, $y_{1} = A sin (k x - ω t), y_{2} = A sin (k x - ω t + ϕ)$ and $y_{3} = A sin (k x - ω t + 2 ϕ)$ are superimposed so that $y_{1} + y_{2} + y_{3} = 0$ at al positions. Then the value of $ϕ$ will be