When two waves of the same amplitude (a0) are superposed, the resultant wave have amplitude √2 a₀. What is the phase difference between the two waves?

CONCEPT:

The equation of wave is given by:

y = A sin (kx - ωt + ϕ)

where x is the position at any time t, A is amplitude, t is time, and ω is the angular frequency.

• Phase: The above equation is said to be in phase (kx-ωt+ϕ). It is an angle representation of the periods spanned by the variable.
• Amplitude (A): maximum displacement from the mean position.
  • When two waves superimpose, the resultant amplitude is found using the parallelogram rule of the vector.

The amplitude of the resultant wave:

\(A=√{A_1^2+A_2^2+2A_1A_2cosϕ}\)

where A1 is the amplitude of the first wave, A2 is the amplitude of the second wave, and ϕ is the phase difference between both the waves.

CALCULATION:

Given that A1 = A2 = a0 and A = √2 a0

We know that  

\(A=√{A_1^2+A_2^2+2A_1A_2cosϕ}\)

\(√2a_0=√{a_0^2+a_0^2+2a_0a_0cosϕ}\)

\(2a_0^2={2a_0^2+2a_0^2cosϕ}\)

\(1={1+cosϕ}\)

\(0={cos{\phi}}\)

φ = \(\dfrac{\pi}{2}\)

So the correct answer is option 1.