Question

An undamped 1.02-kg horizontal spring oscillator has a spring constant of 29.9 N/m. While oscillating, it is found to have a speed of 2.21 m/s as it passes through its equilibrium position. What is its amplitude of oscillation? What is the oscillator\'s total mechanical energy as it passes through a position that is 0.701 of the amplitude away from the equilibrium position?

Answer 1

According to the given problem,

Ek = m ∙ v² / 2
Ek = 1.02*2.21² / 2
Ek = 2.4901 J

Ep = Ek

Ep = k ∙ x² / 2
2.4901 = 29.9 ∙ x² / 2
x = 0.4082 m.

The total mechanical energy remains the same throughout the cycle, so
Em = Epmax = Ekmax = 2.49.01 J

Answer 2

NOTE; Put your question values and solve for unknown variables:

The total mechanical energy of the oscillator at any position equals the sum of its kinetic and potential energies at that position. At the equilibrium position the potential energy is zero, so the total energy equals the kinetic energy, as given by

$$E_{\text{tot}}=\frac{1}{2}m{v}_{\text{max}}^2$$

where E_tot denotes the total energy, m is the oscillator's mass, and v_max is its speed at the equilibrium position (which is the greatest speed it attains during oscillation). Total mechanical energy is conserved by an undamped harmonic oscillator, so the oscillator has this same total mechanical energy, E_tot, at the endpoints of its oscillation. At these positions the oscillator is momentarily at rest and at its maximum amplitude, therefore the kinetic energy is zero and the total energy is equal to the potential energy,

$$E_{\text{tot}}=\frac{1}{2}k{A}^2$$

where k denotes the spring constant and A is the oscillation amplitude. Equate the two expressions for the total mechanical energy that you previously obtained and solve for the amplitude.

$$A=v_{\text{max}}\sqrt{\frac{m}{k}}$$

Substitute the given values to obtain the numerical result.

$$A=\left(3.31\text{ m/s}\right)\sqrt{\frac{2.28\text{ kg}}{28.7\text{ N/m}}}=0.933\text{ m}$$

As mentioned previously, the total mechanical energy is the same at every position, including one that is 0.620$0.620$ of the amplitude away from the equilibrium position. As previously explained, its value is

$$E_{\text{tot}}=\frac{1}{2}m{v}_{\text{max}}^2$$

Enter the given numerical values to find the value of the total mechanical energy.

?tot=12(2.28 kg)(3.31 m/s)2=12.5 J