When the mass is 5.0 cm from its equilibrium point, the percentage of its energy that is kinetic is 75%.
The total energy possessed by the mass under the simple harmonic motion is calculated as follows;
U = ¹/₂kA²
where;
P.E = ¹/₂k(Δx)²
K.E = U - P.E
K.E = ¹/₂kA² - ¹/₂k(Δx)²
[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%.
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