A 0.20-kg object is attached to a spring with spring constant k = 10 N/m and moves with simple harmonic motion over a horizontal frictionless surface.At the instant that it is displaced from equilibrium by -0.050 m,what is its acceleration? A)1 000 m/s² B)-40 m/s² C)0.1 m/s² D)2.5 m/s²

Question

Quizplus · Accepted Answer

The equation for the acceleration of an object in simple harmonic motion is:
a = -ω²x
Where ω is the angular frequency and x is the displacement from equilibrium.

The angular frequency can be found using:
ω = √(k/m)

Plugging in the given values, we get:
ω = √(10 N/m / 0.20 kg) = √50 rad/s

Now we can find the acceleration:
a = -ω²x = -(50 rad/s)²(-0.050 m) = 2.5 m/s²

Therefore, the answer is D: 2.5 m/s².

A 020-Kg Object Is Attached to a Spring with Spring Constant