A particle is in simple harmonic motion along the x axis.The amplitude of the motion is x_m.When it is at x = x₁, its kinetic energy is K = 5 J and its potential energy (measured with U = 0 at x = 0)is U = 3 J.When it is at x = -1/2 x_m, the kinetic and potential energies are: A)K = 6 J and U = 2 J B)K = 6 J and U = -2 J C)K = 8 J and U = 0 J D)K = 0 J and U = 8 J E)K = 0 J and U = -8 J

Question

Quizplus · Accepted Answer

At x = $x_{m}/2$, the total mechanical energy (E) is conserved and equals the sum of kinetic (K) and potential (U) energies. Initially, E = K + U = 5 J + 3 J = 8 J. At $x = -x_{m}/2$, the potential energy is $U = \frac{1}{2}kx^2 = \frac{1}{2}k\left(\frac{x_m}{2}\right)^2 = \frac{1}{4}kx_m^2$. Given that at $x = x_m$, U = 8 J (total energy) - 5 J (kinetic energy at $x_{1/2}x_m$) = 3 J, and U is proportional to $x^2$, at $x = -x_{m}/2$, U = $\frac{1}{4}$ of its value at $x = x_m$, which is $\frac{1}{4} \times 8 J = 2 J$. Thus, K = E - U = 8 J - 2 J = 6 J.

A Particle Is in Simple Harmonic Motion Along the X