Question

What phase difference between two otherwise identical traveling waves, moving in the same direction along a stretched string, will result in the combined wave having an amplitude 1.40 times that of the common amplitude of the two combining waves

Answer 1

`91.14^(\circ)`

Step-by-step explanation:

Wave function form for wave 1

`y_1(x,t)=y_msin(kx-\omega t)`

Wave function form for wave 2

`y_2(x,t)=y_msin(kx-\omega t+\phi)`

Resultant wave function of two waves

`Y(x,t)=y_1(x,t)+y_2(x,t)=y_msin(kx-\omega t)+y_msin(kx-\omega t+\phi)`

`Y(x,t)=y_m(sin(kx-\omega t)+sin(kx-\omega t+\phi))`

`Y(x,t)=2y_mcos(\phi)/(2)sin(kx-\omega t+(\phi)/(2))`

By using the formula

`sinA+sinB=2sin(A+B)/(2)cos(A-B)/(2)`

According to question

`1.4y_m=2y_mcos((\phi)/(2))`

`cos(\phi)/(2)=(1.4)/(2)=0.7`

`(\phi)/(2)=45.57`

`\phi=2* 45.57`

`\phi=91.14^(\circ)`