Question Detail
Question
A particle of mass 0.2 kg undergoes SHM according to the equation: x(t) = 3 sin(δt + δ/4). Where t is in s and x in m. What are the amplitude and the time period of oscillation?
Correct Answer
Option D is the correct answer.
Detailed Explanation
The amplitude is the coefficient of sin, which is 3 m. The angular frequency is δ, but the question has a variable δ. The equation is x = 3 sin(δt + δ/4). The amplitude is 3 m. The time period T = 2π/δ. There is no numerical value for δ given. The options suggest δ = π. If δ = π, T = 2π/π = 2 s.
Hint
The amplitude is the coefficient of sin. The time period T = 2π/ω.