Question

Two waves traveling together along the same line are given by Y1 = 5 sin(rot + 77:/2)
and Y2 = 7 sin (rot + 77:/3). Find (a) the resultant amplitude, (b) the initial phase
angle of the resultant, and (c) the resultant equation of motion.

Answer 1

Answer:

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Answer 2

Answer:

a) The resultant amplitude is, $y_{\max }=11.6$

b) The initial wave angle is given by $&\delta=1.264 \mathrm{rad}$

c) The resultant wave equation is,$y=(11.6)\left[\sin \left(\omega t+72.44^{\circ}\right)\right]$

Explanation:

Given the two waves,

$\begin{aligned}&y_{1}=5 \sin \left(\omega t+\frac{\pi}{2}\right) \\&y_{2}=7 \sin \left(\omega t+\frac{\pi}{3}\right)\end{aligned}$

To find:

(a) the resultant amplitude, (b) the initial phase angle of the resultant, and (c) the resultant equation of motion.

Step 1 of 3:

The resultant is given by, adding to

$&y_{1}=5\left(\sin (\omega t) \cos \left(\frac{\pi}{2}\right)+\cos (\omega t) \sin \left(\frac{\pi}{2}\right)\right)$

$&y_{2}=7\left(\sin (\omega t) \cos \left(\frac{\pi}{3}\right)+\cos (\omega t) \sin \left(\frac{\pi}{3}\right)\right) \\&y=y_{1}+y_{2}$

$&y=\sin (\omega t)\left(5 \cos \left(\frac{\pi}{2}\right)+7 \cos \left(\frac{\pi}{3}\right)\right)+\cos (\omega t)\left(5 \sin \left(\frac{\pi}{2}\right)+7 \sin \left(\frac{\pi}{3}\right)\right)$

$&y=\sin (\omega t)\left(7\left(\frac{1}{2}\right)\right)+\cos (\omega t)\left(5+7\left(\frac{\sqrt{3}}{2}\right)\right) \\&y=3.5 \sin (\omega t)+11.06 \cos (\omega t) \\&y=3.5 \sin (\omega t)+11.06 \cos (\omega t)\left(\frac{\sqrt{3.5^{2}+11.06^{2}}}{\sqrt{3.5^{2}+11.06^{2}}}\right)\end{aligned}$

$&y=\left[\frac{3.5}{\sqrt{3.5^{2}+11.06^{2}}} \sin (\omega t)+\frac{11.06}{\sqrt{3.5^{2}+11.06^{2}}} \cos (\omega t)\right]\left(\sqrt{3.5^{2}+11.06^{2}}\right) \\&y=\left[\frac{3.5}{11.6} \sin (\omega t)+\frac{11.06}{11.6} \cos (\omega t)\right](11.6)$

$&y=\left[\cos \left(72.44^{\circ}\right) \sin (\omega t)+\sin \left(72.44^{\circ}\right) \cos (\omega t)\right](11.6) \\&y=(11.6)\left[\sin \left(\omega t+72.44^{\circ}\right)\right] \quad(3)\end{aligned}$

The resultant wave is given by (3) to be,

$y=(11.6)\left[\sin \left(\omega t+72.44^{\circ}\right)\right]$

Step 2 of 3:

The resultant amplitude is, $y_{\max }=11.6$

The initial wave angle is given by,

$\begin{aligned}&\delta=72.44^{\circ} \\&\delta=72.44\left(\frac{2 \pi}{360^{\circ}}\right) \mathrm{rad} \\&\delta=1.264 \mathrm{rad}\end{aligned}$

Step 3 of 3:

The resultant wave equation is,

$y=(11.6)\left[\sin \left(\omega t+72.44^{\circ}\right)\right]$