The displacement equations of two interfering waves are given by
y₁=10 sin (ω t+π/3) cm, y₂=5[sin ω t+√3 cos ω t] cm respectively.
The amplitude of the resultant wave is _______ cm.

Solution

Physics sir · 4 steps

Rewrite the second wave as one sine first. Then you will see both waves move exactly in step.

1. Combine into one sine. Write y₂ = 5 sin ω t + 5√3 cos ω t in the form A sin(ω t + φ). Its amplitude is A = √(5² + (5√3)²) = √(25 + 75) = 10 cm.
2. Find the phase. Take 10 out as a factor. That leaves 1/2 sin ω t + √3/2 cos ω t = sin ω t cos π/3 + cos ω t sin π/3.

   y₂ = 10 sin(ω t + π/3) cm

3. Compare with y1. y₁ = 10 sin(ω t + π/3) has the same amplitude and the same phase, so the phase difference is 0. With φ = 0, the resultant amplitude √(A₁² + A₂² + 2A₁A₂cos φ) becomes A₁ + A₂.
4. So. The resultant amplitude is 10 + 10 = 20 cm.

The common slip: 15. Taking the factor 5 outside the bracket as the amplitude of y₂ gives 10 + 5 = 15. The true amplitude of y₂ is √(5² + (5√3)²) = 10.

The idea to keep. Before you add two waves, bring each one to the form A sin(ω t + φ). The amplitude of a sin ω t + b cos ω t is √(a² + b²), and then the phase difference decides how they add.

Answer from the NTA final answer key. Solution by PrepWiser.