"A mass of 3.08kg oscillates on the end of a horizontal spring with a period of 0.323s. What acceleration does the mass experience when its displacement is 2.85m to the right?"

1.363 1.08*10^3m/s

To find the acceleration experienced by the mass, we can use the equation for the angular frequency (ω) and the displacement (x).

The angular frequency (ω) is defined as:

ω = 2π / T

Where T is the period of oscillation.

Given that the period (T) is 0.323s, we can find the angular frequency (ω) as:

ω = 2π / 0.323 = 19.475 rad/s

The acceleration (a) of the mass can be calculated using the equation of motion for simple harmonic motion (SHM):

a = -ω^2 * x

Where x is the displacement of the mass.

Given that the displacement (x) is 2.85m to the right, we need to convert this into a negative value since it is in the opposite direction of the equilibrium position. Therefore, x = -2.85m.

Substituting the values in the equation, we have:

a = - (19.475 rad/s)^2 * (-2.85m)
= -943.153 m/s^2

Therefore, the acceleration experienced by the mass when its displacement is 2.85m to the right is -943.153 m/s^2 (directed towards the equilibrium position).