Our analysis thus far has been based on the experimental observation that the oscillation of a spring “looks” sinusoidal. It’s time to look at force and acceleration and to see that Newton’s second law predicts sinusoidal motion.

A motion diagram will help us visualize the object’s acceleration. FIGURE 15.10 shows one cycle of the motion, separating motion to the left and motion to the right to make the diagram clear. As you can see, the object’s velocity is large as it passes through the equilibrium point at $x=0$, but $\vec{v}$ is not changing at that point. Acceleration measures the change of the velocity; hence $\vec{a}=\vec{0}$ at $x=0$.

FIGURE 15.10 Motion diagram of simple harmonic motion. The left and right motions are separated vertically for clarity but really occur along the same line.

In contrast, the velocity is changing rapidly at the turning points. At the right turning point, $\vec{v}$ changes from a right-pointing vector to a left-pointing vector. Thus the acceleration $\vec{a}$ at the right turning point is large and to the left. In one-dimensional motion, the acceleration component $a_x$ has a large negative value at the right turning point. Similarly, the acceleration $\vec{a}$ at the left turning point is large and to the right. Consequently, $a_x$ has a large positive value at the left turning point.

Our motion-diagram analysis suggests that the acceleration $a_x$ is most positive when the displacement is most negative, most negative when the displacement is a maximum, and zero when $x=0$. This is confirmed by taking the derivative of the velocity:

then graphing it.

FIGURE 15.11 shows the position graph that we started with in Figure 15.3 and the corresponding acceleration graph. Comparing the two, you can see that the acceleration graph looks like an upside-down position graph. In fact, because $x=A\text{cos}\omega t$, Equation 15.27 for the acceleration can be written

That is, the acceleration is proportional to the negative of the displacement. The acceleration is, indeed, most positive when the displacement is most negative and is most negative when the displacement is most positive.

FIGURE 15.11 Position and acceleration graphs for an oscillating spring. We’ve chosen ${\varphi }_0=0$.

Recall that the acceleration is related to the net force by Newton’s second law. Consider again our prototype mass on a spring, shown in FIGURE 15.12. This is the simplest possible oscillation, with no distractions due to friction or gravitational forces. We will assume the spring itself to be massless.

FIGURE 15.12 The prototype of simple harmonic motion: a mass oscillating on a horizontal spring without friction.

You learned in Chapter 9 that the spring force is given by Hooke’s law:

The minus sign indicates that the spring force is a restoring force, a force that always points back toward the equilibrium position. If we place the origin of the coordinate system at the equilibrium position, as we’ve done throughout this chapter, then $\text{Δ}x=x$ and Hooke’s law is simply $(F_{\text{Sp}}){}_x=-kx$.

The x-component of Newton’s second law for the object attached to the spring is

Equation 15.30 is easily rearranged to read

You can see that Equation 15.31 is identical to Equation 15.28 if the system oscillates with angular frequency $\omega =\sqrt{k\text{/}m}$. We previously found this expression for $\omega$ from an energy analysis. Our experimental observation that the acceleration is proportional to the negative of the displacement is exactly what Hooke’s law would lead us to expect. That’s the good news.

The bad news is that $a_x$ is not a constant. As the object’s position changes, so does the acceleration. Nearly all of our kinematic tools have been based on constant acceleration. We can’t use those tools to analyze oscillations, so we must go back to the very definition of acceleration:

Acceleration is the second derivative of position with respect to time. If we use this definition in Equation 15.31, it becomes

Equation 15.32, which is called the equation of motion, is a second-order differential equation. Unlike other equations we’ve dealt with, Equation 15.32 cannot be solved by direct integration. We’ll need to take a different approach.