Problem: What is the spring constant of a spring which requires 2.6 seconds to complete an oscillation when a mass of 3 kilograms is placed on it?
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
Solution: We know that the angular frequency of an object in simple harmonic motion is `sqrt(k/m).
The information given is that the object completes a cycle every 2.6 seconds. Thus the object completes 2 `pi radians every 2.6 seconds. This implies a rate of 2 `pi radians/ 2.6 seconds = 2.416 radians/second. This is the angular frequency.
Since we know m, we know that 2.416 radians/second = `sqrt[k/( 3 kg)]. Solving for k we obtain k = 3 kg( 2.416 radians/second) ^ 2 = 17.51 N/m. In symbols, we solve `omega = `sqrt(k/m) for k, obtaining k = m * `omega ^ 2, then substitute the known values of m and the `omega found above.