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 −0.060 m,what is its acceleration? A) 360 m/s² B) −30 m/s² C) 3.0 m/s² D) 0.18 m/s²

Question

Quizplus · Accepted Answer

The equation for the acceleration of an object in simple harmonic motion is given by a = -ω^2x, where ω is the angular frequency (ω = 2πf) and x is the displacement from equilibrium.

First, we need to find the angular frequency:

ω = √(k/m) 
ω = √(10 N/m / 0.20 kg) 
ω = √50 rad/s

Then, we can find the acceleration:

a = -ω^2x 
a = -(50 rad/s)^2(-0.060 m) 
a = 3.0 m/s^2

Therefore, the answer is (C) 3.0 m/s^2.

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