A 0.400-kg object is attached to a spring with spring constant k = 10.0 N/m and moves with simple harmonic motion over a horizontal frictionless surface.At the instant that it is displaced from equilibrium by −0.0550 m,what is its acceleration? A)2 200 m/s² B)−220 m/s² C)0.220 m/s² D)1.38 m/s² E)0.550 m/s²

Question

Quizplus · Accepted Answer

The equation for the acceleration of an object in simple harmonic motion is given by:

a = -ω²x

where ω is the angular frequency (ω = √(k/m)) and x is the displacement from equilibrium.

Plugging in the given values, we have:

ω = √(k/m) = √(10 N/m / 0.400 kg) = 2.236 Ns/m

x = -0.0550 m

a = -ω²x = -(2.236 Ns/m)² (-0.0550 m) = 1.38 m/s²

Therefore, the answer is D.

A 0400-Kg Object Is Attached to a Spring with Spring Constant