Question

An object with mass 3.0 kg is executing simple harmonic motion, attached to a spring with spring constant 290 N/m . When the object is 0.016 m from its equilibrium position, it is moving with a speed of 0.50 m/s .

a) Calculate the amplitude of the motion.

Express your answer to two significant figures and include the appropriate units.

b) Calculate the maximum speed attained by the object.

Express your answer to two significant figures and include the appropriate units.

Answer 1

As total mechanical energy is always conserved in an isolated system.

Total Mechanical energy = Potential Energy + kinetic energy

Now, mechanical energy at 0.016 m from the equilibrium position.

Total Energy = (1/2)*k*x² + (1/2)*M*V²

Total Energy = (0.5)*290*0.016² + (0.5)*3*0.50²

Total energy = 0.037 + 0.375

Total Energy = 0.412 Joules

a) when the speed of object is zero it has the maximum distance from equilibrium position. so, kinetic energy become zero as the object have zero speed.

Total Energy = Potential Energy of object

0.412 = (0.5)*290*X²

solving for X,

X = 0.053 m

thus the amplitude is 0.053 m.

b) the maximum speed of the object is at the equilibrium position. at the equilibrium position it's potential energy is zero.

Total Energy = Kinetic energy

0.412 = (0.5)*3*V²

solving for V,

V = 0.52 m/s

Thus the maximum speed attained by the object is 0.52 m/s.