15.3 Energy in SHM

We’ve begun to develop the mathematical language of simple harmonic motion, but thus far we haven’t included any physics. We’ve made no mention of the mass of the object or the spring constant of the spring. An energy analysis, using the tools of Chapter 10, is a good starting place.

FIGURE 15.8 shows an object oscillating on a spring, our prototype of simple harmonic motion. Now we’ll specify that the object has mass m, the spring has spring constant k, and the motion takes place on a frictionless surface. You learned in Chapter 10 that the elastic potential energy when the object is at position x is $U_{Sp} = \frac{1}{2} k {(Δ x)}^{2}$ , where $Δ x = x - x_{eq}$ is the displacement from the equilibrium position $x_{eq}$ . In this chapter we’ll always use a coordinate system in which $x_{eq} = 0$ , making $Δ x = x$ . There’s no chance for confusion with gravitational potential energy, so we can omit the subscript Sp and write the elastic potential energy as

U = \frac{1}{2} k x^{2}

(15.17)

Thus the mechanical energy of an object oscillating on a spring is

E = K + U = \frac{1}{2} m v^{2} + \frac{1}{2} k x^{2}

(15.18)

The figure illustrates an energy versus displacement graph for a mass oscillating due to a horizontal spring attached to a left wall in simple harmonic motion.

FIGURE 15.8 Full Alternative Text

The lower portion of Figure 15.8 is an energy diagram, showing the parabolic potential-energy curve $U = \frac{1}{2} k x^{2}$ and the kinetic energy $K = E - U$ . Recall that a particle oscillates between the turning points where the total energy line E crosses the potential-energy curve. The left turning point is at $x = - A$ , and the right turning point is at $x = + A$ . To go beyond these points would require a negative kinetic energy, which is physically impossible.

You can see that the particle has purely potential energy at $x = \pm A$ and purely kinetic energy as it passes through the equilibrium point at $x = 0 .$ At maximum displacement, with $x = \pm A$ and $v = 0$ , the energy is

E (at x = \pm A) = U = \frac{1}{2} k A^{2}

(15.19)

At $x = 0$ , where $v = \pm v_{max}$ , the energy is

E (at x = 0) = K = \frac{1}{2} m (v_{max})^{2}

(15.20)

The system’s mechanical energy is conserved because the surface is frictionless and there are no external forces, so the energy at maximum displacement and the energy at maximum speed, Equations 15.19 and 15.20, must be equal. That is

\frac{1}{2} m (v_{max})^{2} = \frac{1}{2} k A^{2}

(15.21)

Thus the maximum speed is related to the amplitude by

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

(15.22)

This is a relationship based on the physics of the situation.

Earlier, using kinematics, we found that

v_{max} = \frac{2 π A}{T} = 2 π f A = ω A

(15.23)

Comparing Equations 15.22 and 15.23, we see that frequency and period of an oscillating spring are determined by the spring constant k and the object’s mass m:

ω = \sqrt{\frac{k}{m}} f = \frac{1}{2 π} \sqrt{\frac{k}{m}} T = 2 π \sqrt{\frac{m}{k}}

(15.24)

These three expressions are really only one equation. They say the same thing, but each expresses it in slightly different terms.

Equations 15.24 tell us that the period and frequency are related to the object’s mass m and the spring constant k. It is perhaps surprising, but the period and frequency do not depend on the amplitude A. A small oscillation and a large oscillation have the same period.