JEE Advance - Physics (2015 - Paper 2 Offline - No. 6)
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 $${I_r} = cA_r^2 = 3c{A^2} = 3{I_0}$$.
Comments (0)
