A 0.250-kg stone is attached to an ideal spring and undergoes simple harmonic oscillations with a period of 0.640 s. What is the force constant (spring constant) of the spring?
A) 12.1 N/m
B) 0.610 N/m
C) 2.45 N/m
D) 24.1 N/m
E) 0.102 N/m

Question

A  0.250-kg  stone is attached to an ideal spring and undergoes simple harmonic oscillations with a period of  0.640 s. What is the force constant (spring constant) of the spring?
A)  12.1 N/m
B)  0.610 N/m
C)  2.45 N/m
D)  24.1 N/m
E)  0.102 N/m

Quizplus · Accepted Answer

The period T of a mass-spring system is given by T = 2π√(m/k), where m is the mass and k is the spring constant. Solving for k gives k = (4π²m)/T². Substituting m = 0.250 kg and T = 0.640 s, we find k ≈ 24.1 N/m.

A 0250-Kg Stone Is Attached to an Ideal Spring and Spring