JEE Advance - Physics (2015 - Paper 2 Offline - No. 6)

Four harmonic waves of equal frequencies and equal intensities I0 have phase angles 0, $${\pi \over 3},{{2\pi } \over 3}$$ and $$\pi$$. When they are superposed, the intensity of the resulting wave is nI0. The value of n is
Answer
3

Explanation

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 $${I_r} = cA_r^2 = 3c{A^2} = 3{I_0}$$.

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