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Q31P

Expert-verifiedFound in: Page 1076

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Add the quantities ${{\mathit{y}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{10}}{\mathbf{sin}}{\mathit{\omega}}{\mathit{t}}$, ${{\mathit{y}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{15}}{\mathbf{sin}}{\left(\omega t+30\xb0\right)}$and ${{\mathit{y}}}_{{\mathbf{3}}}{\mathbf{=}}{\mathbf{5}}{\mathbf{sin}}{\left(\omega t-45\xb0\right)}$ using the phasor method**

The sum of wave is $\left(26.83\right)\mathrm{sin}\left(\omega t+8.5\xb0\right)$.

The equation of first wave is ${y}_{1}=10\mathrm{sin}\omega t$

The equation of second wave is ${y}_{2}=15\mathrm{sin}\left(\omega t+30\xb0\right)$

The equation of third wave is ${y}_{3}=5\mathrm{sin}\left(\omega t-45\xb0\right)$

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

The horizontal component of the resultant wave is given as:

${y}_{h}=10\mathrm{cos}0\xb0+15\mathrm{cos}30\xb0+5c\mathrm{os}\left(-45\xb0\right)\phantom{\rule{0ex}{0ex}}{y}_{h}=10+13+3.54\phantom{\rule{0ex}{0ex}}{y}_{h}=26.54$

The vertical component of the resultant wave is given as:

${y}_{v}=10\mathrm{sin}0\xb0+15\mathrm{sin}30\xb0+5\mathrm{sin}\left(-45\xb0\right)\phantom{\rule{0ex}{0ex}}{y}_{v}=3.96$

The resultant amplitude of waves is given as:

${y}_{r}=\sqrt{{y}_{h}^{2}+{y}_{v}^{2}}\phantom{\rule{0ex}{0ex}}{y}_{r}=\sqrt{{\left(26.54\right)}^{2}+{\left(3.96\right)}^{2}}\phantom{\rule{0ex}{0ex}}{y}_{r}=26.83$

The direction of the resultant wave is given as:

$\begin{array}{rcl}\mathrm{tan}\theta & =& \frac{{y}_{v}}{{y}_{h}}\\ \mathrm{tan}\theta & =& \frac{3.96}{26.54}\\ \theta & =& 8.5\xb0\end{array}$

The sum of the wave is given as:

$y={y}_{r}\mathrm{sin}\left(\omega t+\theta \right)$

Substitute all the values in equation.

$y=\left(26.83\right)\mathrm{sin}\left(\omega t+8.5\xb0\right)$

Therefore, the sum of wave is $\left(26.83\right)\mathrm{sin}\left(\omega t+8.5\xb0\right)$.

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