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  • Three waves,  <img alt="\begin{aligned} & y_1=A \sin (k x-\omega t) \\ & y_2=A \sin (k x-\omega t+\phi) . \\ & y_3=A \sin (k x-\omega t+2 \phi) \end{aligned}" src="https:

Three waves, 

\begin{aligned} & y_1=A \sin (k x-\omega t) \\ & y_2=A \sin (k x-\omega t+\phi) . \\ & y_3=A \sin (k x-\omega t+2 \phi) \end{aligned} 

are superimposed so that y_1+y_2+y_2=0 at all positions. Then, the value of \phi is - 

Option: 1

\frac{2\pi }{3}


Option: 2

\frac{\pi }{3}


Option: 3

\frac{4\pi }{3}


Option: 4

\frac{\pi }{2}


Answers (1)

best_answer

A_1, A_2, A_3 all are equal to A acting at angles shown in figure. We can see that resultant of these three came out tobe zero.

\phi=\frac{2 \pi}{3}=120^{\circ} ; 2 \phi=240^{\circ}

 

Posted by

Devendra Khairwa

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