Question

Three sinusoidal waves of the same frequency travel along a string in the positive $x$ -direcdon. Their amplitudes are $y$ , $\frac{y}{2}$ and $\frac{y}{3}$ and their phase constants are $0$ , $\frac{\pi }{2}$ and $\pi$ respectively. What is the amplitude of the resultant wave?

• A. 0.63 y
• B. 0.72 y
• C. 0.83 y
• D. 0.52 y

Answer 1

Answer: (C)

$Y_1=y\sin (\omega t+0)$
$Y_2=\frac{y}{2}\sin (\omega t+\frac{\pi }{2})$
$Y_3=\frac{y}{3}\sin (\omega t+\pi )$
According to superposition principle
$Y=Y_1+Y_2+Y_3$
$=y\sin \omega t+\frac{y}{2}\cos \omega t+\frac{y}{3}\sin (\omega t+\pi )$
$=y\sin \omega t-\frac{y}{3}\sin \omega t+\frac{y}{2}\cos \omega t$
$=\frac{2y}{3}\sin \omega t+\frac{y}{2}\cos \omega t$
$=y(\frac{2}{3}\sin \omega t+\frac{1}{2}\cos \omega t)$
$=\frac{y}{6}(4\sin \omega t+3\cos \omega t)$
$Y=\frac{4y}{6}\sin \omega t+\frac{3y}{6}\cos \omega t$
$A_1=\frac{2y}{3}$ and $A_2=\frac{y}{2}$
$A=\sqrt{A_1^2+A_2^2}$
$=\sqrt{\frac{4y^2}{9}+\frac{y^2}{4}}$
$=y\sqrt{\frac{16+9}{36}}=\frac{y}{6}\times 5$
$=\frac{5y}{6}=0.83y$