Question

Add the quantities ${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{sin}\mathit{\omega }\mathit{t}$, ${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{15}\mathbf{sin}\left(\omega t+30°\right)$and${\mathit{y}}_{\mathbf{3}}\mathbf{=}\mathbf{5}\mathbf{sin}\left(\omega t-45°\right)$ using the phasor method

Short answer

The sum of wave is $\left(26.83\right)\sin \left(\omega t+8.5°\right)$.

Step-by-step solution

Identification of given data

The equation of first wave is$y_1=10\sin \omega t$

The equation of second wave is $y_2=15\sin \left(\omega t+30°\right)$

The equation of third wave is $y_3=5\sin \left(\omega t-45°\right)$

Understanding the concept

The amplitude of the resultant wave is equal to the vector sum of the amplitude of each wave.

Determination of vertical and horizontal component of resultant wave

The horizontal component of the resultant wave is given as:

$$\begin{gathered} y_h=10\cos 0°+15\cos 30°+5c\mathrm{os}\left(-45°\right) \\ {y}_h=10+13+3.54 \\ {y}_h=26.54 \end{gathered}$$

The vertical component of the resultant wave is given as:

$$\begin{gathered} y_v=10\sin 0°+15\sin 30°+5\sin \left(-45°\right) \\ {y}_v=3.96 \end{gathered}$$

The resultant amplitude of waves is given as:

$$\begin{gathered} y_r=\sqrt{y_h^2+y_v^2} \\ {y}_r=\sqrt{{\left(26.54\right)}^2+{\left(3.96\right)}^2} \\ {y}_r=26.83 \end{gathered}$$

The direction of the resultant wave is given as:

$\begin{array}{rcl}\tan \theta & =& \frac{y_v}{y_h}\\ \tan \theta & =& \frac{3.96}{26.54}\\ \theta & =& 8.5°\end{array}$

Determination of sum of wave

The sum of the wave is given as:

$y=y_r\sin \left(\omega t+\theta \right)$

Substitute all the values in equation.

$y=\left(26.83\right)\sin \left(\omega t+8.5°\right)$

Therefore, the sum of wave is $\left(26.83\right)\sin \left(\omega t+8.5°\right)$.