A mass m at the end of a spring vibrates with a frequency of 0.87 Hz. When an additional 602 g mass is added to m, the frequency is 0.59 Hz. What is the value of m?

Answer 1

[1/(2 pi)]*(k/m)^1/2 = 0.87

[1/(2 pi)]*[k/(m+602)]^1/2 = 0.59

Take the ratios of the two equations
[(m+602)/m]^1/2 = 1.4746

(m+602)/m = 2.174
602/m = 1.174
m = 513 g