QJEEAdv15Phb348: JEE Advanced 2015 Physics - Class 11 JEEAdv15Phb348 Solution: Four harmonic waves of equal frequencies and equal intensities…

The intensity of a wave is proportional to the square of its amplitude i.e., I₀ = cA², where c is a constant. The amplitudes of four harmonic waves are equal as their intensities are equal. Let these waves be travelling along the x direction with wave vector k and angular frequency $\omega$ . The resultant displacement of these waves is given by

$y = {y_1} + {y_2} + {y_3} + {y_4}$

$= A\sin (\omega t - kx + 0) + A\sin (\omega t - kx + \pi /3) + Asin(\omega t - kx + 2\pi /3) + A\sin (\omega t - kx + \pi )$

$= A\sin (\omega t - kx + \pi /3) + Asin(\omega t - kx + 2\pi /3)$

$= 2A\sin (\omega t - kx + \pi /2)\cos (\pi /6)$

$= \sqrt 3 A\cos (\omega t - kx)$ .

The amplitude of the resultant wave is ${A_r} = \sqrt 3 A$ and its intensity is ${I_r} = cA_r^2 = 3c{A^2} = 3{I_0}$ .