A mass m at the end of a spring oscillates with a frequency of 0.89 hz. when an additional 603 g mass is added to m, the frequency is 0.63 hz. what is the value of m?

The solution for this problem:

Given:

f1 = 0.89 Hz

f2 = 0.63 Hz

Δm = m2 - m1 = 0.603 kg

The frequency of mass-spring oscillation is:

f = (1/2π) √ (k/m)

k = m (2πf) ²

Then we know that k is constant for both trials, we have:

k = k

m1 (2πf1) ² = m2 (2πf2) ²

m1 = m2 (f2/f1) ²

m1 = (m1+Δm) (f2/f1) ²

m1 = Δm / ((f1/f2) ²-1)

m 1 = 0.603 / (0.89/0.63) ^2 - 1

= 0.609 kg or 0.61kg or 610 g