Question

When a mass is attached to a vertical spring, the spring is stretched a distance d. The mass is then pulled down from this position and released. It undergoes 52 oscillations in 28.3 s. What was the distance d?

Answer 1

F = m g = k d
so
k/m = g/d

but w = 2 pi f = sqrt (k/m)
but given f = 52/28.3
so
2 pi * 52/28.3 = sqrt (k/m)
so
k/m = 133
so
9.8/d = 133
d = .0737 meters