The total mechanical energy $$E$$ is equal to the maximum potential energy, which occurs at the amplitude, $$x=A$$. So, $$E = U(A)$$. We have $$8.0 \text{ J} = (2.0 \text{ J/m}^2)A^2$$. Solving for $$A^2$$ gives $$A^2 = 8.0/2.0 = 4.0 \text{ m}^2$$. Taking the square root gives the amplitude $$A = 2.0$$ m.

The total mechanical energy is determined by the maximum gravitational potential energy. The maximum height of the center of mass above its equilibrium position is $$h = d(1-\cos\theta_{max})$$. The total energy is $$E = Mgh = Mgd(1-\cos\theta_{max})$$. To increase $$E$$, one must increase $$M$$, $$g$$, $$d$$, or $$\theta_{max}$$. Increasing the mass $$M$$ while other factors are constant will increase the total energy.

The total mechanical energy is conserved. At the maximum angular displacement, the bob is momentarily at rest, so its kinetic energy is zero. The total energy is equal to the gravitational potential energy at that point. The height of the bob above the lowest point is $$h = L - L\cos\theta_{max} = L(1 - \cos\theta_{max})$$. Therefore, the total energy is $$E = mgh = mgL(1 - \cos\theta_{max})$$.

The total energy is $$E = K + U$$. We want the position where $$K=U$$. This means $$E = U + U = 2U$$. The potential energy is $$U = \frac{1}{2}kx^2$$ and the total energy is $$E = \frac{1}{2}kA^2$$. Substituting these into $$E = 2U$$ gives $$\frac{1}{2}kA^2 = 2(\frac{1}{2}kx^2)$$, which simplifies to $$A^2 = 2x^2$$. Solving for $$x$$ gives $$x = \pm \frac{A}{\sqrt{2}}$$.

The velocity is $$v(t) = \frac{dx}{dt} = -A\omega\sin(\omega t)$$. The kinetic energy is $$K(t) = \frac{1}{2}mv(t)^2 = \frac{1}{2}m(-A\omega\sin(\omega t))^2 = \frac{1}{2}mA^2\omega^2\sin^2(\omega t)$$. The total energy is $$E = \frac{1}{2}kA^2 = \frac{1}{2}m\omega^2A^2$$. Substituting $$E$$ into the expression for $$K(t)$$ gives $$K(t) = E\sin^2(\omega t)$$.

This question tests AP Physics C: Mechanics understanding of energy conservation in simple harmonic oscillators, focusing on the mathematical expression. In SHM without friction, total mechanical energy remains constant throughout oscillation, expressed as the sum of kinetic and potential energies equaling the maximum potential energy at amplitude. Choice A correctly states ½mv² + ½kx² = ½kA², showing that K + U equals the total energy E. Choice B incorrectly uses subtraction instead of addition. Choice C omits the ½ factors. Choice D rearranges incorrectly, suggesting potential energy equals total plus kinetic. Choice E incorrectly states total energy is zero. To help students: derive the energy conservation equation from first principles, practice identifying correct forms of conservation laws, and verify equations using dimensional analysis and limiting cases.

This question tests AP Physics C: Mechanics understanding of energy transformations in simple harmonic oscillators, specifically finding where kinetic and potential energies are equal. When K = U, each equals half the total energy since E = K + U. Setting U = E/2: ½kx² = ½(½kA²), which simplifies to x² = A²/2, giving |x| = A/√2 ≈ 0.707A. With A = 0.10 m, |x| = 0.10/√2 ≈ 0.0707 m ≈ 0.071 m. Choice B is correct because it matches this calculation. Choice C (0.050 m = A/2) is a common error from assuming linear rather than quadratic energy relationships. To help students: emphasize that energy varies as x², practice solving for positions where K and U have specific ratios, and use energy bar charts to visualize energy distribution at different positions.

The total mechanical energy of a simple harmonic oscillator is constant. At the point of maximum displacement (the amplitude $$A$$), the object is momentarily at rest, so its kinetic energy is zero. At this point, all the energy is stored as potential energy in the spring, which is given by $$U_s = \frac{1}{2}kA^2$$. Therefore, the total mechanical energy is $$E = \frac{1}{2}kA^2$$.

The total mechanical energy of a mass-spring system in simple harmonic motion is given by $$E = \frac{1}{2}kA^2$$, where $$k$$ is the spring constant and $$A$$ is the amplitude. Since the energy is proportional to the square of the amplitude, tripling the amplitude ($$A_{new} = 3A_{orig}$$) results in the new energy being $$3^2 = 9$$ times the original energy.

The total energy is $$E = \frac{1}{2}kA^2 = \frac{1}{2}(8.0)(0.50)^2 = 1.0$$ J. At any position $$x$$, the total energy is the sum of kinetic and potential energy: $$E = \frac{1}{2}mv^2 + \frac{1}{2}kx^2$$. At $$x=0.30$$ m, $$1.0 = \frac{1}{2}(2.0)v^2 + \frac{1}{2}(8.0)(0.30)^2$$. This simplifies to $$1.0 = v^2 + 0.36$$, so $$v^2 = 0.64$$, and $$v = 0.80$$ m/s.