15.1.1 Kinematics of Simple Harmonic Motion

We’ll start by describing simple harmonic motion mathematically—that is, with kinematics. Then in Section 15.4 we’ll take up the dynamics of how forces cause simple harmonic motion.

FIGURE 15.3 shows a SHM position graph—such as the one generated by the air-track glider—in its “normal” position. For the moment we’ll assume that the oscillator starts at maximum displacement $(x = + A)$ at $t = 0$ . Also shown is the oscillator’s velocity-versus-time graph, which we can deduce from the slope of the position graph.

The figure consists of two parts that illustrate a position versus time and velocity versus time graphs respectively.

FIGURE 15.3 Full Alternative Text

The instantaneous velocity is zero at the instants when $x = \pm A$ because the slope of the position graph is zero. These are the turning points in the motion.
The position graph has maximum slope when $x = 0$ , so these are points of maximum speed. When $x = 0$ with a positive slope—maximum speed to the right—the instantaneous velocity is $v_{x} = + v_{max}$ , where $v_{max}$ is the amplitude of the velocity curve. Similarly, $v_{x} = - v_{max}$ when $x = 0$ with a negative slope—maximum speed to the left.

Although these are empirical observations (we don’t yet have any “theory” of oscillation), we can see that the position graph, with a maximum at $t = 0$ , is a cosine function with amplitude A and period T. We can write this as

x (t) = A cos (\frac{2 π t}{T})

(15.2)

where the notation $x (t)$ indicates that x is a function of time t.

NOTE We have been using the trigonometry of right triangles to find the components of vectors. In this chapter, we begin to use the more general idea that sine and cosine are oscillatory functions of the angle. Appendix A reminds you what graphs of sin and cos look like. Don’t forget that the arguments of sine and cosine functions are in radians; this will be true throughout our study of oscillations and waves. Be sure to set your calculator to radian mode before doing calculations.

Because $cos (0 rad) = c o s (2 π rad) = 1$ , we can see that $x = A$ at $t = 0$ and again at $t = T$ . In other words, this is a cosine function with amplitude A and period T. Notice that x passes through zero at $t = \frac{1}{4} T$ and $t = \frac{3}{4} T$ because $cos (\frac{1}{2} π) = cos (\frac{3}{2} π) = 0$ .

We can write Equation 15.2 in two alternative forms. First, because the oscillation frequency is $f = 1 / T$ , we can write

x (t) = A cos (2 π f t)

(15.3)

Second, recall from Chapter 4 that a particle in circular motion has angular velocity $ω$ that is related to the period by $ω = 2 π / T$ , where $ω$ is in rad/s. For oscillations and waves, $ω$ is called the angular frequency. We can write the position in terms of $ω$ as

x (t) = A cos (ω t)

(15.4)

Most of our work with oscillations and waves will be in terms of the angular frequency.

Now that we’ve defined $f = 1 / T$ , we see that $ω$ , f, and T are related by

ω (in rad / s) = \frac{2 π}{T} = 2 π f (in Hz)

(15.5)

Be careful! Both f and $ω$ are frequencies, but they’re not the same and they’re not interchangeable. Frequency f is the true frequency, in cycles per second, and it’s always measured in Hz. Angular frequency $ω$ is useful because it’s related to the angle of the cosine function and, as you’ll learn in the next section, to a circular-motion analog of SHM, but it’s always in rad/s.

Just as the position graph is a cosine function, you can see that the velocity graph in Figure 15.3 is an “upside-down” sine function with the same period. We can write the velocity function as

v_{x} (t) = - v_{max} sin (\frac{2 π t}{T})

(15.6)

where the minus sign inverts the graph. This function is zero at $t = 0$ and again at $t = T$ .

A photograph of a quartz tuning fork. — This 3-mm-long quartz tuning fork is the timing element of a digital watch. It is cut from a perfect quartz crystal to give it a very well-defined oscillation frequency. Quartz is a *piezoelectric material*, which means that a stress or deformation creates an electric field in the crystal. The crystal tuning fork is part of a circuit that responds to the oscillating electric field as the tuning fork vibrates, generating a signal at that frequency, and at the same time feeds the oscillating voltage back to the crystal to keep the vibration going. The quartz crystal in a watch is designed to oscillate at $32,768 Hz = 2^{15} Hz$ because digital circuits can easily divide by powers of 2 to generate a timing pulse exactly once a second.

NOTE $v_{max}$ is the maximum speed and thus is a positive number.

We deduced Equation 15.6 from the experimental results, but we could equally well find it from the position function of Equation 15.2. After all, velocity is the time derivative of position. TABLE 15.2 reminds you of the derivatives of the sine and cosine functions. Using the derivative of the position function, we find

v_{x} (t) = \frac{d x}{d t} = - \frac{2 π A}{T} sin (\frac{2 π t}{T}) = - 2 π f A sin (2 π f t) = - ω A sin ω t

(15.7)

Comparing Equation 15.7, the mathematical definition of velocity, to Equation 15.6, the empirical description, we see that the maximum speed of an oscillation is

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

(15.8)

Not surprisingly, the object has a greater maximum speed if you stretch the spring farther and give the oscillation a larger amplitude.

TABLE 15.2 Derivatives of sine and cosine functions

\frac{d}{d t} (a sin (b t + c)) = + a b cos (b t + c)

\frac{d}{d t} (a cos (b t + c)) = - a b sin (b t + c)

EXAMPLE 15.1 A system in simple harmonic motion

An air-track glider is attached to a spring, pulled 20.0 cm to the right, and released at $t = 0 s$ . It makes 15 oscillations in 10.0 s.

What is the period of oscillation?
What is the object’s maximum speed?
What are the position and velocity at $t = 0.800 s ?$

MODEL An object oscillating on a spring is in SHM.

SOLVE

The oscillation frequency is

$f = \frac{15 oscillations}{10 .0 s} = 1.50 oscillations/s = 1.50 Hz$

Thus the period is $T = 1 / f = 0.667 s$ .
The oscillation amplitude is $A = 0.200 m$ . Thus

$v_{max} = \frac{2 π A}{T} = \frac{2 π (0.200 m)}{0.667 s} = 1.88 m / s$
The object is at $x = + A$ at $t = 0 s$ . This is exactly the oscillation described by Equations 15.2 and 15.6. The position at $t = 0.800 s$ is

$\begin{array}{l} x & = & A cos (\frac{2 π t}{T}) = (0.200 m) cos (\frac{2 π (0.800 s)}{0.667 s}) \\ = & (0.200 m) cos (7.54 rad) = 0.0625 m = 6.25 cm \end{array}$

The velocity at this instant of time is

$\begin{array}{l} v_{x} & = & - v_{max} sin (\frac{2 π t}{T}) = - (1.88 m/s) sin (\frac{2 π (0.800 s)}{0.667 s}) \\ = & - (1.88 m/s) sin (7.54 rad) = - 1.79 m/s = - 179 cm/s \end{array}$

At $t = 0.800 s,$ which is slightly more than one period, the object is 6.25 cm to the right of equilibrium and moving to the left at $179 cm / s$ . Notice the use of radians in the calculations.

EXAMPLE 15.2 Finding the time

A mass oscillating in simple harmonic motion starts at $x = A$ and has period T. At what time, as a fraction of T, does the object first pass through $x = \frac{1}{2} A ?$

SOLVE Figure 15.3 showed that the object passes through the equilibrium position $x = 0$ at $t = \frac{1}{4} T$ . This is one-quarter of the total distance in one-quarter of a period. You might expect it to take $\frac{1}{8} T$ to reach $\frac{1}{2} A$ , but this is not the case because the SHM graph is not linear between $x = A$ and $x = 0$ . We need to use $x (t) = A cos (2 π t / T)$ . First, we write the equation with $x = \frac{1}{2} A :$

x = \frac{A}{2} = A cos (\frac{2 π t}{T})

Then we solve for the time at which this position is reached:

t = \frac{T}{2 π} {cos}^{- 1} (\frac{1}{2}) = \frac{T}{2 π} \frac{π}{3} = \frac{1}{6} T

REVIEW The motion is slow at the beginning and then speeds up, so it takes longer to move from $x = A$ to $x = \frac{1}{2} A$ than it does to move from $x = \frac{1}{2} A$ to $x = 0$ . Notice that the answer is independent of the amplitude A.