Four waves are to be sent along the same string, in the same direction: y_(1)(x,t)=(4.90mm)sin(2pi x-430pi t+0.42pi ) y_(2)(x,t)=(4.90mm)sin(2pi x-430pi t+0.93pi ) y_(3)(x,t)=(4.90mm)sin(2pi x-430pi t+1.42pi ) y_(4)(x,t)=(4.90mm)sin(2pi x-430pi t+1.93pi ) What is the amplitude of the resultant wave?

Here's how to find the amplitude of the resultant wave:<br /><br />**1. Understanding the Problem**<br /><br />We have four waves with the same amplitude, frequency, and wavelength traveling in the same direction. The only difference between them is their phase. To find the resultant wave's amplitude, we need to add these waves together, taking into account their phase differences.<br /><br />**2. Phasor Representation**<br /><br />The easiest way to add waves with different phases is to use phasor representation. Each wave can be represented as a vector (phasor) in the complex plane. The length of the vector is the amplitude of the wave, and the angle it makes with the positive x-axis is the phase.<br /><br />**3. Adding the Phasors**<br /><br />* **Pairs:** Notice that the phase difference between $y_1$ and $y_2$ is the same as the phase difference between $y_3$ and $y_4$ (0.51π). This means these pairs will add to form two resultant waves with the same amplitude.<br /><br />* **Calculate the amplitude of one pair:** Let's consider $y_1$ and $y_2$. The amplitude of the resultant of these two waves can be found using the formula for adding two phasors:<br /><br /> $A_{12} = \sqrt{A^2 + A^2 + 2A^2 cos(\phi_2 - \phi_1)}$<br /><br /> Where A is the amplitude of the individual waves (4.90 mm), and $\phi_2 - \phi_1$ is the phase difference (0.51π).<br /><br /> $A_{12} = \sqrt{2(4.90)^2 + 2(4.90)^2 cos(0.51\pi)} = \sqrt{2(4.90)^2(1 + cos(0.51\pi))} \approx 4.90\sqrt{2(1 - 0.0314)} \approx 6.87 mm$<br /><br />* **Adding the resultant phasors:** Since the two resultant waves ($y_{12}$ and $y_{34}$) also have the same amplitude ($A_{12} = A_{34}$) and their phase difference is also 0.51π (1.42π - 0.93π), we can use the same formula again:<br /><br /> $A_{total} = \sqrt{A_{12}^2 + A_{34}^2 + 2A_{12}A_{34} cos(0.51\pi)}$<br /><br /> $A_{total} = \sqrt{2(6.87)^2 + 2(6.87)^2 cos(0.51\pi)} \approx 6.87\sqrt{2(1 - 0.0314)} \approx 9.68 mm$<br /><br />**4. Final Answer**<br /><br />The amplitude of the resultant wave is approximately 9.68 mm.<br />