Waves Part 1

Given, equation of wave of first string, $y_{1}=A_{1} \sin k(x-v t)$ Equation of wave of second string, $y_{2}=A_{2} \sin k(x-v t+x_{0})$ $A_{1}=12 \text{mm}, A_{2}=5 \text{mm}, x_{0}=3.5 \text{cm}$ and $k=6.28 \text{cm}^{-1}$ The equation of waves after substitution of values can beexpressed as $y_{1}=12 \sin 6.28(x-v t)$ and $y_{2}=5 \sin 6.28(x-v t+3.5)$ The phase difference between two waves can be calculated as $\Delta \phi = \frac{2 \pi}{\lambda} \Delta x$ From equation of waves, the value of $\Delta x=(x-v t+3.5)-(x-v t)=3.5 \text{cm}$ We know that, wave number is k = 2 π / λ. Now, the phase difference is ∆ϕ = k∆x= 6.28 × 3.5 = 2 π × 3.5 = 7 π The amplitude of resulting wave can be calculated by using thefollowing relation, $A_{\text{net}}=\sqrt{A_{1}^{2}+A_{2}^{2}+2 A_{1} A_{2} \cos \Delta \phi}$ Substituting the values in above expression, $A_{\text{net}}=\sqrt{12^{2}+5^{2}+2 \times 12 \times 5 \times \cos 7 \pi}$ $=\sqrt{144+25+120 \times (-1)}$ $A_{\text{net}}=\sqrt{49}=7 \text{mm}$ Thus, the amplitude of resultant wave is 7 mm.