A 4.50-kg mass hangs vertically downward from a massless spring of spring constant 147.0 N/m.

Question

A 4.50-kg mass hangs vertically downward from a massless spring of spring constant 147.0 N/m.

What is the length that the spring stretches from its equilibrium position when the mass is hung from the spring?

The system is set into a simple harmonic motion by pulling the mass down 0.123 m and releasing it. What is the frequency of the motion?



What is the amplitude of the oscillation?

Answer 1

Extension = Mg/K =4.5*9.8/K = (4.5*9.8 / 147.0 ) m = 0.3m

Frequency = 1/2pi(SqrtK/M) =5.72/6.28 = 0.91 Hz

Amplitude = 0.123 m