Question

A particle is in simple harmonic motion along the x-axis with an amplitude of xm. 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 its potential energy is 8 J, it is at:

A. x = 0
B. x = x1
C. x = xm/2
D. x = xm/2
E. x = xm

Answer 1

When the potential energy of the particle is 8 J in simple harmonic motion, the particle is at its maximum displacement from equilibrium, denoted as x = xm.

Step-by-step explanation:

The question involves understanding the energy transformation in simple harmonic motion (SHM), specifically the oscillation of kinetic and potential energy as an object moves along the x-axis. Because the total energy is conserved in a system undergoing SHM, knowing the potential energy at any point allows for the determination of its position. In this scenario, the given potential energies 3 J and 8 J must correspond to points along the oscillation where the total mechanical energy (sum of kinetic and potential) remains constant.

When the potential energy is given as 3 J, and the kinetic energy is 5 J, the total mechanical energy is 8 J. As the potential energy increases to 8 J, the kinetic energy becomes zero because the total energy is conserved. This scenario occurs only at the turning points where the amplitude is maximum, ±xm. Therefore, when the potential energy of the particle is 8 J, the particle is at x = xm, which corresponds to the maximum displacement from equilibrium.