Question

Six harmonic waves of equal frequency and equal intensity $I_0$ have phase angles $0,\frac{\pi }{6},\frac{2\pi }{6},...\frac{5\pi }{6}$ . When they are superimposed, the intensity of the resulting wave is $n{I}_0$ Find the value of $3n$ (nearly). $[\text{Given}\frac{1}{{\sin }^{2}15{}^{\circ }}\simeq 15]$

Answer 1

It form a sum of $6$ vector

$\therefore O{O}^{\prime }=\frac{A_0\sin \left(\frac{n\theta }{2}\right)}{\sin \left(\frac{\theta }{2}\right)}$

$\Rightarrow =O{O}^{\prime }=\frac{A_0\sin (6\times \frac{30{}^{\circ }}{2})}{\sin 15{}^{\circ }}=\frac{A_0}{\sin 15{}^{\circ }}$

$\therefore I\propto (A_r{)}^2\propto \frac{A_0^2}{{\sin }^2{15}^{\circ }}=I_{0}14.9\approx 15I_0$

As $I=n{I}_0=15I_0\therefore n=15$

$\therefore 3n=45$