NTA Abhyas JEE Main2020PhysicsWaves and SoundPractice
The equations of three waves are given by y 1 = A 0 sin k x − ω t , y 2 = 3 2 A 0 sin k x − ω t + ϕ and y 3 = 4 A 0 cos k x − ω t . Theses waves are in the same direction and are superimposed. The phase difference between the resultant-wave and the first wave is π 4 and ϕ = π n ≤ π 2 , then what is the value of n ?
Correct answer
12.00
Step-by-step solution
tan π 4 = BC AC = A 0 ( 4 + 3 2 sin ϕ ) A 0 ( 1 + 3 2 cos ϕ ) ⇒ cos ϕ - sin ϕ = 1 2 Squaring both sides, ⇒ cos 2 ϕ + sin 2 ϕ - 2 cos ϕ sin ϕ = 1 2 ⇒ 2 sin ϕ cos ϕ = 1 2 ⇒ sin 2 ϕ = 1 2 ⇒ ϕ = 1 2 sin - 1 1 2 = π 12