A mass of 0.40 kg,attached to a spring with a spring constant of 80 N/m,is set into simple harmonic motion.What is the magnitude of the acceleration of the mass when at its maximum displacement of 0.10 m from the equilibrium position? A)zero B)5 m/s² C)10 m/s² D)20 m/s²

Question

Quizplus · Accepted Answer

At maximum displacement, the mass is at the extreme end of its motion, so it has zero velocity. The acceleration of the mass at maximum displacement can be calculated using the equation:
a = -ω^2x
where ω is the angular frequency of the motion, and x is the displacement from equilibrium. The angular frequency can be found from the spring constant k and the mass m:
ω = sqrt(k/m)
Substituting the given values, we get:
ω = sqrt(80 N/m / 0.40 kg) = 10 rad/s
At x = 0.10 m, the acceleration is:
a = -(10 rad/s)^2 (0.10 m) = -10 m/s^2
The magnitude of the acceleration is positive, so we take the absolute value:
|a| = 10 m/s^2
Therefore, the correct choice is D, with an acceleration magnitude of 20 m/s^2.

A Mass of 0