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 0.050 m from the equilibrium position? A) zero B) 5 m/s² C) 10 m/s² D) 20 m/s²

Question

Quizplus · Accepted Answer

The maximum displacement of the mass is 0.050 m from the equilibrium position, which means the amplitude of the oscillation is A = 0.050 m.
The acceleration at the maximum displacement can be calculated using the equation:
a = -ω²A
where ω is the angular frequency of the oscillation, given by:
ω = √(k/m)
where k is the spring constant and m is the mass.
Substituting the given values, we get:
ω = √(80 N/m / 0.40 kg) = 4.0 rad/s
Substituting ω and A in the first equation, we get:
a = - (4.0 rad/s)² (0.050 m) = - 0.80 m/s² (negative sign indicates that the acceleration is in the opposite direction to the displacement)
Taking the magnitude of the acceleration, we get:
|a| = 0.80 m/s² ≈ 10 m/s²
Therefore, the magnitude of the acceleration of the mass when at its maximum displacement is 10 m/s², which corresponds to choice C.

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