A particle is subjected to two simple harmonic motions in the same direction having . . .

Question : A particle is subjected to two simple harmonic motions in the same direction having equal magnitude and equal frequency. If the resultant amplitude is equal to the amplitude of the individual motion. Find the phase difference between two individual motions.

Doubt by Dev

Solution : 

Let us represent the two SHM by following two equations of motion

x₁=a₁cosωt
x₂=a₂cos(ωt+Φ)

where
x₁ & x₂ are the displacement of the particle during the two SHM at any time t 
a₁ & a₂ are the amplitude of the particles during two SHM
Φ is the constant phase difference between the two SHM

According to superposition principle 
x=x₁+x₂

x=a₁cosωt+a₂cos(ωt+Φ)
[∵ cos(A+B) = cosA cosB - sinA sinB ]

x=a₁cosωt+a₂(cosωt cosΦ - sinωt sinΦ)
x=a₁cosωt+a₂cosωt cosΦ - a₂sinωt sinΦ
x=(a₁+a₂cosΦ) cosωt - (a₂sinΦ) sinωt

Let 
a₁+a₂cosΦ = A cosθ --------(1)
a₂sinΦ = A sinθ --------(2)

where A is the resultant amplitude and θ is the resultant phase angle w.r.t. first SHM.

x = A cosθ cosωt - A sinθ sinωt
x = A[cosθ cosωt - sinθ sinωt]
x = A[cos(θ+ωt)]
[∵ cosA cosB - sinA sinB = cos(A+B)]

x = A[cos(ωt+θ)]

Now, squaring and adding eq (1) and (2)

(a₁+a₂cosΦ)² + (a₂sinΦ)² = (A cosθ)²  + (A sinθ)² 
a₁²+a₂²cos²Φ+2a1a2cosΦ + a₂²sin²Φ = A²cos²θ + A²sin²θ
a₁²+a₂²(sin²Φ+cos²Φ)+2a₁a₂cosΦ = A²(sin²θ+cos²θ)

[∵ sin²θ+cos²θ=1]

a₁²+a₂²+2a₁a₂cosΦ = A²
A² = a₁²+a₂²+2a₁a₂cosΦ
A = √(a₁²+a₂²+2a₁a²cosΦ)

According to Question 

A = a1 = a2 = a (let)
a = √(a2+a2+2×a×acosΦ )
a2 = 2a2+2a2cosΦ 
a2 = 2a2(1+cosΦ)
1 = 2 (1+cosΦ)
1/2 = 1+cosΦ 
1/2-1 =cosΦ 
-1/2 = cosΦ 
cos2π/3 = cosΦ 
Φ = 2π/3 radian.
Therefore, required phase difference is 2π/3 radian.