two sinusoidal waves y1(x,t), y2(x,t) have the same wavelength and travel together in the same direction…

Explanation:

Step1: Recall the formula for the amplitude of the resultant wave

The formula for the amplitude $A$ of the resultant of two - sinusoidal waves with amplitudes $A_1$ and $A_2$ and phase difference $\Delta\phi$ is $A=\sqrt{A_1^{2}+A_2^{2}+2A_1A_2\cos\Delta\phi}$. Here, $A_1 = 4.0\times10^{- 3}\text{m}$, $A_2 = 3.0\times10^{-3}\text{m}$, and $\Delta\phi=\frac{\pi}{3}$.

Step2: Calculate the cosine of the phase - difference

$\cos\Delta\phi=\cos\frac{\pi}{3}=\frac{1}{2}$.

Step3: Substitute the values into the amplitude formula

[ \begin{align*} A&=\sqrt{(4.0\times 10^{-3})^{2}+(3.0\times 10^{-3})^{2}+2\times(4.0\times 10^{-3})\times(3.0\times 10^{-3})\times\frac{1}{2}}\ &=\sqrt{16\times10^{-6}+9\times10^{-6}+12\times10^{-6}}\ &=\sqrt{(16 + 9+12)\times10^{-6}}\ &=\sqrt{37\times10^{-6}}\ &\approx6.08\times10^{-3}\text{m}=6.08\text{mm} \end{align*} ]

Step4: Find the phase constant of the resultant wave

We can use the formula $\tan\theta=\frac{A_2\sin\Delta\phi}{A_1 + A_2\cos\Delta\phi}$. $\sin\Delta\phi=\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}$, $A_1 = 4.0\times10^{-3}\text{m}$, $A_2 = 3.0\times10^{-3}\text{m}$, and $\cos\Delta\phi=\frac{1}{2}$. [ \begin{align*} \tan\theta&=\frac{3.0\times10^{-3}\times\frac{\sqrt{3}}{2}}{4.0\times10^{-3}+3.0\times10^{-3}\times\frac{1}{2}}\ &=\frac{3\sqrt{3}/2}{4 + 3/2}\ &=\frac{3\sqrt{3}/2}{11/2}\ &=\frac{3\sqrt{3}}{11}\approx0.477 \end{align*} ] $\theta=\arctan(0.477)\approx0.44\text{rad}\approx25.2^{\circ}$

Step5: Write the resultant wave in standard form

The standard form of a sinusoidal wave is $y(x,t)=A\sin(kx-\omega t+\theta)$. Substituting $A = 6.08\times10^{-3}\text{m}$ and $\theta\approx0.44\text{rad}$, we get $y(x,t)=6.08\times10^{-3}\sin(kx-\omega t + 0.44)$ (in SI units).

Answer:

(a) The amplitude of the resultant wave is approximately $6.08\text{mm}$. (b) The phase constant of the resultant wave is approximately $0.44\text{rad}$ or $25.2^{\circ}$. (c) The resultant wave in standard form is $y(x,t)=6.08\times10^{-3}\sin(kx-\omega t + 0.44)$ (in SI units).