A spring stretches by 0.018 m when a 3.6-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 2.9 Hz?

Answer 1

k=mg/x=3.6•9.8/0.018 =1960
f=2.9 Hz => ω=2πf=2π•2.9 = 18.22 rad/s
ω²=k/M => M=k/ ω²=1960/(18.22)² = 5.9 kg
Δm=M-m=5.9 – 3.6 – 2.3 kg

Answer 2

You do not need to find the difference. The second to last step is the answer.