A mass of 0.250 kg is attached to a spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the force 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 the equation:

T = 2π√(m/k)

where T is the period, m is the mass, and k is the force constant of the spring.

Rearranging the equation, we get:

k = (4π^2m) / T^2

Plugging in the given values, we get:

k = (4π^2 * 0.250 kg) / (0.640 s)^2

k = 24.1 N/m

Therefore, the force constant of the spring is 24.1 N/m, which corresponds to choice C.

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