Physics

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?

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1. 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

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posted by drwls

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