Physics
posted by BJ on .
A mass m at the end of a spring vibrates with a frequency of 0.93 Hz. When an additional 607 g mass is added to m, the frequency is 0.63 Hz. What is the value of m?

Frequency = [1/(2 pi)] sqrt (k/m)
0.93 = [1/(2 pi)] sqrt (k/m)
0.63 = [1/(2 pi)] sqrt [k/(m+0.607)]
Take the ratio. The k will cancel out.
0.93/0.63 = 1.476 = sqrt[(m+0.607)/m]
= sqrt (1 + 0.607/m )
2.179 = 1 + 0.607/m
1.170 = .607/m
Solve for m. It will be in kg since I wrote the added mass in kg