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 spring constant of the spring?

Question

Quizplus · Accepted Answer

The frequency of oscillation can be calculated as:
f = 1/T
where T is the period of oscillation.

Substituting the given values:
f = 1/0.640 s
f = 1.5625 Hz

The angular frequency, ω can be calculated as:
ω = 2πf

Substituting the value of f:
ω = 2π × 1.5625 rad/s
ω = 9.817 rad/s

The spring constant, k can be calculated as:
k = mω^2
where m is the mass of the stone.

Substituting the given values:
k = 0.250 kg × (9.817 rad/s)^2
k = 24.103 N/m

Therefore, the spring constant of the spring is 24.1 N/m, which is closest to option C (24.1 N/m).

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