A0.22 kg mass at the end of a spring oscillates 2.9 times per second with an amplitude of 0.13 m .

1. determine the speed when it passes the equilibrium point.

2. determine the speed when it is 0.12 m from equilibrium.

3. determine the total energy of the system.

2.3687599 m/s

0.91106 m/s

0.617213012 J

Explanation:

f = Frequency =

A = Amplitude = 0.13 m

k = Spring constant

m = Mass of object = 0.22 kg

Angular velocity is given by

Velocity is given by

Speed when it passes the equilibrium point is 2.3687599 m/s

Frequency is given by

x = Displacement = 0.12 m

In this system the energies are conserved

The speed when it is 0.12 m from equilibrium is 0.91106 m/s

The energy in the system is given by

The total energy of the system is 0.617213012 J

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