Question

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?

Answer 1

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