Post a New Question

Physics

posted by .

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?

  • Physics -

    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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question