Question

An object attached to the end of a spring slides on a horizontal frictionless surface with simple harmonic motion. Assume the maximum displacement of the object is A/2 so that it oscillates between +A/2 and −A/2. Determine the location(s) L at which the object's kinetic energy and potential energy are the same in terms of A. Enter your value(s) as a comma-separated list.

Answer 1

Locations L where an object's kinetic and potential energies are the same during SHM can be found by equating the kinetic and potential energy equations and solving for displacement x. These locations would be at displacements between 0 and ±A/2. Without specific values for mass m and spring constant k, numerical solution for L cannot be provided.

Step-by-step explanation:

To determine the locations (L) where the object's kinetic energy and potential energy are the same during simple harmonic motion (SHM), we must understand the energy transformation in SHM. In undamped SHM, the total mechanical energy (E) is constant and is the sum of kinetic (K) and potential (U) energies. The potential energy U is greatest when the spring is at maximal compression or extension, which would be at ±A/2 in this case, and it is zero at the equilibrium position (x = 0).

Using the conservation of energy:

E = K + U

Since the total energy is conserved, at the points where K = U, each will be half of the total energy E. To find L, we calculate this balance point using:

K = ½ U, which occurs at the displacement where U = ½ kx² = ½ mv² = K.

Ignoring the constants and focusing on the proportions:

x² = v²

At the equilibrium position (x = 0), the speed is maximal, and as the object moves away from equilibrium, speed decreases while displacement x increases until it reaches ±A/2. At some point before reaching ±A/2, the kinetic and potential energies will be equal. The specific points L would be where x is neither 0 nor ±A/2; these values can be calculated by setting the potential energy equation equal to the kinetic energy equation and solving for x.

Therefore, to find the values of L in terms of A which satisfy K = U, we would typically use the known values for k (spring constant) and m (mass) and solve this equation:

kx² = mv²

However, without numerical values for k and m, we cannot provide a specific numerical answer for L. Typically, L would be a value between 0 and ±A/2 where the spring's displacement squared is half of the maximum potential energy.