Answer :
Answer:
[tex]91.14^{\circ}[/tex]
Explanation:
Wave function form for wave 1
[tex]y_1(x,t)=y_msin(kx-\omega t)[/tex]
Wave function form for wave 2
[tex]y_2(x,t)=y_msin(kx-\omega t+\phi)[/tex]
Resultant wave function of two waves
[tex]Y(x,t)=y_1(x,t)+y_2(x,t)=y_msin(kx-\omega t)+y_msin(kx-\omega t+\phi)[/tex]
[tex]Y(x,t)=y_m(sin(kx-\omega t)+sin(kx-\omega t+\phi))[/tex]
[tex]Y(x,t)=2y_mcos\frac{\phi}{2}sin(kx-\omega t+\frac{\phi}{2})[/tex]
By using the formula
[tex]sinA+sinB=2sin\frac{A+B}{2}cos\frac{A-B}{2}[/tex]
According to question
[tex]1.4y_m=2y_mcos(\frac{\phi}{2})[/tex]
[tex]cos\frac{\phi}{2}=\frac{1.4}{2}=0.7[/tex]
[tex]\frac{\phi}{2}=45.57[/tex]
[tex]\phi=2\times 45.57[/tex]
[tex]\phi=91.14^{\circ}[/tex]