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A mass attached to a spring oscillates in simple harmonic motion with an amplitude of 10 cm. When the mass is 5.0 cm from its equilibrium point, what percentage of its energy is kinetic

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onyebuchinnaji

When the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

Total energy of the mass

The total energy possessed by the mass under the simple harmonic motion  is calculated as follows;

U = ¹/₂kA²

where;

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

Potential energy of the mass at 5 cm from equilibrium point

P.E = ¹/₂k(Δx)²

Kinetic energy of mass

K.E = U - P.E

K.E = ¹/₂kA² - ¹/₂k(Δx)²

Percentage of its energy that is kinetic

[tex]K.E (\%) = \frac{U - P.E}{U} \times 100\%\\\\K.E (\%) =\frac{\frac{1}{2}kA^2 - \frac{1}{2}k(\Delta x)^2 }{\frac{1}{2}kA^2} \times 100\%\\\\K.E (\%) = \frac{A^2 - (\Delta x)^2}{A^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - (10-5)^2}{10^2} \times 100\%\\\\K.E (\%) = \frac{10^2 - 5^2}{10^2} \times 100\%\\\\K.E (\%) = 75\%[/tex]

Thus, when the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.

Learn more about kinetic energy here: https://brainly.com/question/25959744

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