Question

What is the maximum elastic potential energy for simple horizontal spring oscillator whose equation of motion is given by x=0.35 m sin(7rad/)t?

Answer 1

The maximum elastic potential energy for simple horizontal spring is 2.7 J.

Maximum elastic potential energy of the spring

The maximum elastic potential energy of the spring is calculated as follows;

U = ¹/₂kA²

where;

• A is the amplitude of the wave
• k is the spring constant

k = ω²m

where;

• m is mass given as 0.9 kg
• ω is angular speed given as 7 rad/s

k = (7²)(0.9)

k = 44.1 N/m

U = 0.5(44.1)(0.35²)

U = 2.7 J

Thus, the maximum elastic potential energy for simple horizontal spring is 2.7 J.

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The complete question is below:

The mass on the end of the spring is 0.900kg.