An object that weighs 2.450 N is attached to an ideal massless spring and undergoes simple harmonic oscillations with a period of 0.640 s.What is the spring constant of the spring?
A) 2.45 N/m
B) 12.1 N/m
C) 24.1 N/m
D) 0.102 N/m
E) 0.610 N/m

Question

Quizplus · Accepted Answer

The period of simple harmonic motion is given by T = 2π√(m/k), where m is the mass attached to the spring and k is the spring constant.

Rearranging the formula, we get: k = (4π^2m) / T^2

Substituting the given values, we get: k = (4π^2 x 2.450) / 0.640^2 = 24.1 N/m

Therefore, the spring constant of the spring is 24.1 N/m. Option C is the correct choice.

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