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a 0.50-kg mass is attached to a spring of spring constant 20 n/m along a horizontal, frictionless surface. the object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. what is the amplitude of vibration?

Sagot :

The vibration amplitude of a spring with a constant of 20 N/m which has a speed of 1.5 m/s with a mass of 0.50 kg which is associated = 0.237 m.

Oscillation Energy

When an object oscillates on a spring, the kinetic energy and potential energy of the mass-spring system change with time.

The total energy (the kinetic energy + potential energy) is constant.

The kinetic energy of an object moving with speed v is:

½ mv².

The potential energy of a spring with a constant k is:

= ½ kx².

The total energy = ½ kx² + ½ mv² = ½ kA².

The total energy equation gives the general property of simple harmonic oscillations which is directly proportional to the square of the amplitude.

And, we have:

m = 0.50 kg

k = 20 N/m

v = 1.5 m/s

Remember, the total energy = ½ kA².

So,

A² = (m)(v)² / k

= (0.50) (1.5)² / 20

= 0.05625

A = [tex]\sqrt{0.05625}[/tex]

= 0.237 m

Learn more about oscillation energy here: https://brainly.com/question/28543819

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