Question

A glider of mass m is attached to a spring that has a force constant k. The glider is set into simple harmonic motion with amplitude a. Determine the positions (by magnitude) at which the kinetic and the potential energies of the glider are equal. Use a coordinate system whose center is in the equilibrium position of the glider, where the x-axis is directed to the right, and the point of the spring's attachment to the wall has a negative x-coordinate. Express your answer in terms of the variables m, k, and a.

Options:
a. ±a/√2
b. ±a/2
c. ±a
d. ±a√2

Answer 1

In simple harmonic motion, the positions where the kinetic and potential energies of the glider are equal occur at ±a/√2, where a is the amplitude of motion.

Step-by-step explanation:

To determine the positions at which the kinetic and potential energies of the glider are equal, we need to consider the properties of simple harmonic motion (SHM). In SHM, the total mechanical energy of the system is conserved and can be expressed as the sum of kinetic energy (KE) and potential energy (PE). At the equilibrium position (x = 0), the potential energy is at its minimum (zero), and the kinetic energy is at its maximum. Conversely, at the amplitude positions (x = ±a), the kinetic energy is zero, and the potential energy is at its maximum. The potential energy stored in the spring at any position x is given by U(x) = 1/2kx².To find the positions where kinetic energy equals potential energy, we set KE = PE. Since the total mechanical energy E is conserved and is equal to the maximum potential energy, which occurs at the amplitude, E = U(max) = 1/2ka². Because E = KE + PE and at the points we are looking for KE = PE, we can write E = 2KE. Thus, 1/2ka² = 2(1/2mv²) → mv² = 1/4ka². Using the relationship between the maximum speed (vmax) and the amplitude for SHM, which is vmax = ωa where ω is the angular frequency, ω = √k/m, we obtain m(√k/ma)² = 1/4ka². Simplifying this gives a²/2 = a²/4, which is only true if x² = a²/2 → x = ±a/√2.