Question

Find the sum y of the following quantities: ${\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{10}\mathbf{}\mathbf{sin}\mathbf{}\mathit{\omega }\mathit{t}$ and ${\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{sin}\left(\omega t+30°\right)$

Short answer

The sum of wave is $\left(17\right)\sin \left(\omega t+13.3°\right)$.

Step-by-step solution

Identification of given data

The amplitude of first wave is $A_1=10$

The amplitude of second wave is $A_2=8$

The phase difference for both waves is $\varphi =30°$

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

Determination of resultant amplitude and direction of resultant amplitude of wave

The resultant amplitude of wave is given as:

$A=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \varphi }$

Substitute all the values in equation.

$\begin{array}{rcl}A& =& \sqrt{{\left(10\right)}^2+{\left(8\right)}^2+2\left(10\right)\left(8\right)\left(\cos 30°\right)}\\ A& =& \sqrt{302.56}\\ A& \approx & 17\end{array}$

The direction of the resultant amplitude is given as:

role="math" localid="1663050213428" $\tan \theta =\frac{A_2\sin \varphi }{A_1+A_2\cos \varphi }$

Substitute all the values in equation.

$\begin{array}{rcl}\tan \theta & =& \frac{\left(8\right)\left(\sin 30°\right)}{10+\left(8\right)\left(\cos 30°\right)}\\ \theta & =& 13.3°\end{array}$

Determination of sum of wave

The sum of the wave is given as:

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

Substitute all the values in equation.

$y=\left(17\right)\sin \left(\omega t+13.3°\right)$

Therefore, the sum of wave is $\left(17\right)\sin \left(\omega t+13.3°\right)$.