Suppose that the period of a particular ideal mass-spring system is 5 s. What would be the period of the system if the mass were two times the original mass?

Answer: T2 = 7.07s

Explanation: The period of a loaded spring of spring constant k and mass m is given by

T = 2π √m/k

With 2π constant and k, it can be seen with little algebra that

T² is proportional to mass m

Hence (T1) ²/m1 = (T2) ²/m2

Where T1 = 5, T2 = ?, let m1 = m hence m2 = 2m.

By substituting, we have that

5²/m = (T2) ²/2m

25 / m = (T2) ²/2m

25 * 2m = (T2) ² * m

25 * 2 = (T2) ²

50 = (T2) ²

T2 = √50

T2 = 7.07s