These two waves travel along the same string:

$$\begin{gathered} {\mathbf{y}}_{\mathbf{1}}\left(x,t\right)\mathbf{=}\left(4.60\mathrm{mm}\right)\mathbf{sin}\left(2\mathrm{\pi x}-400\mathrm{\pi t}\right) \\ {\mathbf{y}}_{\mathbf{2}}\left(x,t\right)\mathbf{=}\left(5.60\mathrm{mm}\right)\mathbf{sin}\left(2\mathrm{\pi x}-400\mathrm{\pi t}+0.80\mathrm{\pi rad}\right) \end{gathered}$$

What is the amplitude (a) and (b) what is the phase angle (relative to wave 1) of the resultant wave? (c) If a third wave of amplitude 5.00 mmis also to be sent along the string in the same direction as the first two waves, what should be its phase angle in order to maximize the amplitude of the new resultant wave?

$$\begin{gathered} {\mathrm{y}}_1\left(\mathrm{x},\mathrm{t}\right)=\left(4.60\mathrm{mm}\right)\sin \left(2\mathrm{\pi x}-400\mathrm{\pi t}\right) \\ {\mathrm{y}}_1\left(\mathrm{x},\mathrm{t}\right)=\left(5.6\mathrm{mm}\right)\sin \left(2\mathrm{\pi x}-400\mathrm{\pi t}+0.80\mathrm{\pi }\mathrm{rad}\right) \end{gathered}$$

Amplitude of the third wave,$y_{3m}=5.00\mathrm{mm}$

When two waves travel alongthesame string, the amplitude of each wave can be found relative to the resultant wave. Usingthephasor technique, the phase angle can be calculated.

Formula:

The phase angle of two connecting lines, $\mathrm{\theta }=\frac{\mathrm{Adjacent}\mathrm{Side}}{\mathrm{Resultant}}......\left(1\right)$

From the given equations of the waves, we can see that the amplitude of the first wave is 4.6 m and the amplitude of the second wave is 5.6 mm. From the wave equations, it is also clear that the phase difference between these two waves is$0.80\pi \mathrm{rad}\mathrm{or}144°$. This means that the angle between the amplitude of these two waves is $0.80\pi \mathrm{rad}\mathrm{or}144°$So, if we add these two vectors, we will get the resultant. It can be done as follows:

To find the amplitude relative to the first wave with phase difference $144°$, we can use the standard technique for addition of two vectors. The resultant vector will have only y component caused by $y_2\mathrm{as}{\mathrm{y}}_1$has only x component.

$$\begin{gathered} y_R=\left(y_2\right)\sin \left(144°\right)\left(\mathrm{where}{\mathrm{y}}_2=5.6\mathrm{mm}\right) \\ =3.29\mathrm{mm} \end{gathered}$$

Hence, the value of the amplitude is 3.29 mm

x component of the resultant amplitude would be

$$\begin{gathered} y_{Rx}=y_1-y_2\cos \left(180-144\right) \\ =4.6-5.6\cos 36 \\ =0.06950 \end{gathered}$$

To find the angle relative to wave 1, we can use equation (1) in the given figure to the phase of resultant wave,

$$\begin{gathered} \mathrm{\theta }={\cos }^{-1}\left(\frac{0.06950}{3.29}\right) \\ =88.8° \end{gathered}$$

Therefore, the phase angle relative to wave 1 is$88.8°$

For the amplitude to be maximum,the third wave should be in phase with the resultant wave.

So the angle must be equal to $88.8°$relative to the first wave.