A simple motion is represented by: y = 5(sin(3πt) + √3cos(3πt)) cm. The amplitude and time period of the motion are:

The given equation is y = 5(sin(3πt) + √3cos(3πt)).
We can rewrite this equation in the form Rsin(ωt + α), where R is the amplitude and ω is the angular frequency.
Let's use the trigonometric identity: Rsin(ωt + α) = Rsin(ωt)cos(α) + Rcos(ωt)sin(α)
Comparing this with the given equation, we have:
Rcos(α) = 5
Rsin(α) = 5√3
Dividing the second equation by the first, we get:
tan(α) = √3
Therefore, α = π/3
Squaring and adding the two equations, we get:
R²cos²(α) + R²sin²(α) = 5² + (5√3)²
R²(cos²(α) + sin²(α)) = 25 + 75
R² = 100
R = 10 cm (amplitude)
The angular frequency is ω = 3π
The time period T is given by:
T = 2π/ω = 2π/(3π) = 2/3 s
Therefore, the amplitude is 10 cm and the time period is 2/3 s. However, none of the options match this solution. There must be an error in the options provided. The correct answer should be 10 cm, 2/3 s.