With this demonstration you can illustrate the dynamics and energetics of a simple harmonic oscillator.

The page for demonstration 40.12 – Mass-springs with different spring constants and masses, gives the mathematics that describes the motion of a simple harmonic oscillator, specifically a mass hanging from a spring. The equation that describes the forces (gravity and the spring tension) acting on the mass yields the second-order differential equation

m(d²x/dt²) + kx = 0.

where m is the mass hanging from the spring, x is the position of the mass along the axis of oscillation, and k is the spring constant, which is a measure of the stiffness of the spring. Solution of this equation leads to the following equation for the position of the mass along the axis of oscillation, as a function of time:

x = A cos (ωt - φ)

A is the amplitude of the oscillation, which corresponds to the maximum displacement of the mass from its equilibrium position to either turning point. ω (= √(k/m)) is the vibrational frequency in radians per second, and φ is a phase factor, which depends on the initial speed and position of the mass, as does A. (Whether the function one obtains is a sine or a cosine function depends on the choice of initial conditions, that is, whether x₀, the position at t = 0, is the equilibrium position or a turning point.) The period of oscillation is T = 2π/ω, and the frequency in cycles per second is ω/(2π).

When the mass and spring are not oscillating, the scale reads their combined weight, which we may call mg. If you pull down on the mass, the scale reads mg + kΔx, where Δx is the distance by which you have pulled the mass down from its equilibrium position and thus further stretched the spring. If you raise the mass, the scale reads mg - kΔx, where Δx is the distance by which you have raised the mass above its equilibrium position and thus relaxed the spring. As the mass-spring oscillates, then, we expect the scale reading to oscillate accordingly, reaching a maximum at the bottom turning point and a minimum at the top turning point.

We can also see this if we examine the equation for the acceleration of the mass. For completeness, we will include the equation for the velocity of the mass, as well. If we take x =  A cos (ωt - φ), then v = dx/dt = -ωA sin (ωt - φ), and a = d²x/dt² = -ω²A cos (ωt - φ). These equations show us that at the extrema (maximum and minimum x), the velocity is zero, and the acceleration is at its maximum (or minimum), and that at the equilibrium position (x = 0), the acceleration is zero and the velocity is at its maximum (or minimum). Where the acceleration is maximum (or minimum), the force on the mass-spring must also be a maximum (or minimum). At the turning points, then, when the mass is at the bottom or top of its travel, the scale reading is greatest or least, respectively. (At the bottom turning point, the acceleration is maximum upward, and at the top turning point, it is maximum downward. Thus, the scale, which has to provide the force required for this acceleration in addition to mg, registers the greatest force when the mass reaches the bottom, and the least force when it reaches the top.) At the equilibrium position, where the velocity is greatest, the acceleration equals zero, and the force on the mass-spring is mg, the same as when the mass-spring was not oscillating. As the mass oscillates, the scale reading varies sinusoidally, going from the equilibrium reading (mg) down to the minimum reading at the top turning point, rising through mg at the equilibrium position to the maximum reading at the bottom turning point, then decreasing again to mg at the equilibrium position. The highest and lowest scale readings are equidistant above and below what the scale reads when the mass is at the equilibrium position, and their values depend on how far down you pull the mass (or how far up you raise it) to set the sytem oscillating.

We can also understand this by considering the energy of the mass-spring system. This is the sum of its potential energy and its kinetic energy. The potential energy is the combination of gravitational potential energy and the energy stored in the spring. At the top turning point, the gravitational potential energy is at its maximum, and the energy in the stretched spring is at its minimum. At the bottom turning point, the gravitational potential energy is at its minimum, and the energy in the stretched spring is at its maximum. At the equilibrium position, their sum is at its minimum, and displacing the mass from there in either direction increases the potential energy by (1/2)kx². We can thus take the equilibrium position of the system as the zero reference for the potential energy. The kinetic energy is (1/2)mv². The total energy (the sum of potential energy and kinetic energy) must remain constant, except for frictional losses. The equations above, for x and v, show that at the turning points, the potential energy is at its maximum, and the kinetic energy equals zero. At the equilibrium position, the potential energy equals zero, and the kinetic energy is at its maximum.

As the mass descends from the equilibrium position to the bottom turning point, its kinetic energy is converted into potential energy in the spring (while the gravitational potential energy decreases) until, at the turning point, the potential energy is a maximum. The greater the potential energy stored in the spring, the greater the force it exerts on the scale, so as the mass descends from the equilibrium position to the bottom turning point, the scale reading increases from mg to the maximum reading. As the mass returns from the bottom turning point, and the potential energy is converted back into kinetic energy, the force exerted by the spring on the scale decreases, and the scale reading decreases, until the mass reaches the equilibrium position. At this point the potential energy equals zero and the scale reads mg. As the mass continues to rise from the turning point, the potential energy in the spring decreases, but the gravitational potential energy increases. Thus the force exerted by the spring on the scale decreases, and the scale reading decreases. At the top turning point, the gravitational potential energy is at its maximum, but the potential energy in the spring is at its minimum. So the force exerted by the spring on the scale at this point is at its minimum, and the scale reading reaches the minimum value.Gravity then accelerates the mass downward. As the mass descends, it loses gravitational potential energy, and the potential energy in the spring begins to increase. The tension in the spring thus increases, and the scale reading increases. When the mass reaches the equilibrium position, the tension in the spring again equals mg. As noted above, through the cycle, the scale reading varies sinusoidally, going from the equilibrium reading (mg) down to the minimum reading at the top turning point, rising through mg at the equilibrium position to the maximum reading at the bottom turning point, then decreaseing again to mg at the equilibrium position. Also as noted above, the highest and lowest scale readings are equidistant above and below what the scale reads when the mass is at the equilibrium position, and their values depend on how far down you pull the mass (or how far up you raise it) to set the sytem oscillating.