91Ó°ÊÓ

i) The wave equation is,$y(x,t)=y_{m}sin(kx-\mathrm{Ó\neg }t+\mathrm{Ï•})$

ii) Three waves have the same frequency, $f$.

iii) The amplitudes of three waves,$y_1,y_1/2,y_1/3$

iv) Phase constants of three waves, $0,\mathrm{Ï€}/2,\mathrm{Ï€}$

We can find the amplitude by using the phasor and taking the resultant of the given amplitudes. The phase constant of the resultant wave can be calculated from the x and y component of the resultant amplitude. The plot can be drawn by forming the wave equation for the resultant wave.

Phasor diagram for the given waves:

Let y be the resultant of the three given waves.

The horizontal component of y is,

$$\begin{gathered} y_x=y_1-\frac{y_1}{3} \\ =\frac{2}{3}{y}_1 \end{gathered}$$

The vertical component of y is,

$y_y=\frac{y_1}{2}$

Therefore amplitude of the resultant wave is given as:

$$\begin{gathered} y_m=\sqrt{y_x^2+y_y^2} \\ =\sqrt{{\left(\frac{2}{3}{y}_1\right)}^2+{\left(\frac{y_1}{2}\right)}^2} \\ =\sqrt{\frac{4}{9}{y}_1^2+\frac{1}{4}{y}_1^2} \\ =0.83y_1 \end{gathered}$$

Therefore, the value of the amplitude of the resultant wave is $0.83y_1$.

The phase constant of the resultant wave is given as:

$\mathrm{Ï•}={\tan }^{-1}\left(\frac{y_y}{y_x}\right)$

Substitute the values in the above expression, and we get

$$\begin{gathered} \mathrm{ϕ}={\tan }^{-1}\left(\frac{{\displaystyle \frac{y_1}{2}}}{{\displaystyle \frac{2}{3}}{y}_1}\right) \\ \mathrm{ϕ}=\left(\frac{3}{4}\right) \\ =36.87 \\ ~37° \end{gathered}$$

Therefore, the phase constant of the resultant wave is $37°$.

The general equation of the wave is,

$y=y_m\sin \left(kx-\mathrm{Ó\neg }t+\mathrm{Ï•}\right)$

Therefore, the equation of the resultant wave is,

localid="1660985063034" $y=0.83y_{1}sin(kx-\mathrm{Ó\neg }t+{37}^{°})$

The plot of the wavelocalid="1660985065802" $y=0.83y_{1}sin(kx-\mathrm{Ó\neg }t+{37}^{°})$at t=0

Hence, as t increases, the wave is moving in the positive x direction with wave number k and angular velocity $\mathrm{Ó\neg }$.