A particle executes simple harmonic motion represented by displacement function as
x(t) = A sin(ωt + φ)
If the position and velocity of the particle at t = 0 s are 2 cm and 2ω cm s⁻¹ respectively, then its amplitude is x√2 cm where the value of x is _________________.

Solution

Physics sir · 3 steps

Use the starting position and velocity together, and the phase drops out when you square and add.

1. Position at start. Put t = 0 in x = A sin(ω t + φ). This gives A sin φ = 2.
2. Velocity at start. Differentiate to get v = Aω cos(ω t + φ). At t = 0 this is Aω cos φ = 2ω, so A cos φ = 2.

   A² = (A sin φ)² + (A cos φ)² = 2² + 2² = 8

3. So. A = √8 = 2√2 cm. Comparing with x√2 cm gives x = 2.

The common slip: 2.83. Entering the full amplitude 2√2 ≈ 2.83 gives 2.83, not 2. The question asks only for the number in front of √2.

The idea to keep. Given the position and velocity at any instant, the amplitude is A = √(x² + v²/ω²). You never need to find the phase first.

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