"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).