Mass on a Spring

Consider a compact mass $m$ which slides over a frictionless horizontal surface. Suppose that the mass is attached to one end of a light horizontal spring whose other end is anchored in an immovable wall. See Figure 1. At time $t$, let $x(t)$ be the extension of the spring: i.e., the difference between the spring's actual length and its unstretched length. Obviously, $x(t)$ can also be used as a coordinate to determine the instantaneous horizontal displacement of the mass.

Figure 1: Mass on a spring

The equilibrium state of the system corresponds to the situation in which the mass is at rest, and the spring is unextended (i.e., $x=\dot{x}=0$, where $\dot{}\equiv d/dt$). In this state, zero horizontal force acts on the mass, and so there is no reason for it to start to move. However, if the system is perturbed from its equilibrium state (i.e., if the mass is displaced, so that the spring becomes extended) then the mass experiences a horizontal restoring force given by Hooke's law:

$$f (x)= -k\,x.$$ (1)

Here, $k>0$ is the so-called force constant of the spring. The negative sign indicates that $f(x)$ is indeed a restoring force (i.e., if the displacement is positive then the force is negative, and vice versa). Note that the magnitude of the restoring force is directly proportional to the displacement of the mass from its equilibrium position (i.e., $\vert f\vert\propto x$). Of course, Hooke's law only holds for relatively small spring extensions. Hence, the displacement of the mass cannot be made too large. Incidentally, the motion of this particular dynamical system is representative of the motion of a wide variety of mechanical systems when they are slightly disturbed from a stable equilibrium state (see Section 2.4).

Newton's second law of motion leads to the following time evolution equation for the system:

$$m\,\ddot{x} = - k\,x,$$ (2)

where $\ddot{} \equiv d^2/dt^2$. This differential equation is known as the simple harmonic oscillator equation, and its solution has been known for centuries. In fact, the solution is

$$x(t) = a\,\cos(\omega\,t-\phi),$$ (3)

where $a>0$, $\omega>0$, and $\phi$ are constants. We can demonstrate that Equation (3) is indeed a solution of Equation (2) by direct substitution. Plugging the right-hand side of (3) into Equation (2), and recalling from standard calculus that $d(\cos\theta)/d\theta = -\sin\theta$ and $d (\sin\theta)/d\theta = \cos\theta$, so that $\dot{x} = -\omega\,a\,\sin(\omega\,t-\phi)$ and $\ddot{x} = -\omega^2\,a\,\cos(\omega\,t-\phi)$, where use has been made of the chain rule,

$$\frac{d}{dx}\left(f\left[g(x)\right]\right)\equiv \frac{df}{dg}\,\frac{dg}{dx},$$ (4)

we obtain

$$-m\,\omega^2\,a\,\cos(\omega\,t-\phi) =- k\,a\,\cos(\omega\,t-\phi).$$ (5)

It follows that Equation (3) is the correct solution provided

$$\omega = \sqrt{\frac{k}{m}}.$$ (6)

Figure 2: Simple harmonic oscillation.

Figure 2 shows a graph of $x$ versus $t$ obtained from Equation (3). The type of behavior displayed here is called simple harmonic oscillation. It can be seen that the displacement $x$ oscillates between $x=-a$ and $x=+a$. Here, $a$ is termed the amplitude of the oscillation. Moreover, the motion is repetitive in time (i.e., it repeats exactly after a certain time period has elapsed). In fact, the repetition period is

$$T = \frac{2\pi}{\omega}.$$ (7)

This result is easily obtained from Equation (3) by noting that $\cos\theta$ is a periodic function of $\theta$ with period $2\pi$: i.e., $\cos(\theta+2\pi) \equiv \cos\theta$. It follows that the motion repeats every time $\omega\,t$ increases by $2\pi$: i.e., every time $t$ increases by $2\pi/\omega$. The frequency of the motion (i.e., the number of oscillations completed per second) is

$$f = \frac{1}{T} = \frac{\omega}{2\pi}.$$ (8)

It can be seen that $\omega$ is the motion's angular frequency; i.e., the frequency $f$ converted into radians per second. Of course, $f$ is measured in Hertz--otherwise known as cycles per second. Finally, the phase angle, $\phi$, determines the times at which the oscillation attains its maximum displacement, $x=a$. In fact, since the maxima of $\cos\theta$ occur at $\theta=n\,2\pi$, where $n$ is an arbitrary integer, the times of maximum displacement are

$$t_{\rm max} = T \left(n + \frac{\phi}{2\pi}\right).$$ (9)

So, varying the phase angle simply shifts the pattern of oscillation backward and forward in time. See Figure 3.

Figure 3: Simple harmonic oscillation. The solid, short-dashed, and long dashed-curves correspond to $\phi = 0$, $+\pi /4$, and $-\pi /4$, respectively.

Table 1: Simple harmonic oscillation.

$\omega\,t-\phi$	$0$	$\pi/2$	$\pi$	$3\pi/2$
$x$	$+a$	0	$-a$	0
$\dot{x}$	0	$-\omega\,a$	0	$+\omega\,a$
$\ddot{x}$	$-\omega^2\,a$	0	$+\omega^2\,a$	0

Table 1 lists the displacement, velocity, and acceleration of the mass at various different phases of the simple harmonic oscillation cycle. The information contained in this table can easily be derived from Equation (3). Note that all of the non-zero values shown in this table represent either the maximum or the minimum value taken by the quantity in question during the oscillation cycle.

We have seen that when a mass on a spring is disturbed it executes simple harmonic oscillation about its equilibrium position. In physical terms, if the mass's initial displacement is positive ($x>0$) then the restoring force is negative, and pulls the mass toward the equilibrium point ($x=0$). However, when the mass reaches this point it is moving, and its inertia thus carries it onward, so that it acquires a negative displacement ($x<0$). The restoring force then becomes positive, and again pulls the mass toward the equilibrium point. However, inertia again carries it past this point, and the mass acquires a positive displacement. The motion subsequently repeats itself ad infinitum. The angular frequency of the oscillation is determined by the spring stiffness, $k$, and the system inertia, $m$, via Equation (6). On the other hand, the amplitude and phase angle of the oscillation are determined by the initial conditions. To be more exact, suppose that the instantaneous displacement and velocity of the mass at $t=0$ are $x_0$ and $v_0$, respectively. It follows from Equation (3) that

$$x_0 = x(t=0) = a\,\cos\phi,$$ (10)

$$v_0 = \dot{x}(t=0) =a\,\omega\,\sin\phi.$$ (11)

Here, use has been made of the trigonometric identities $\cos(-\theta) \equiv\cos\theta$ and $\sin(-\theta) \equiv-\sin\theta$. Hence, we deduce that

$$a = \sqrt{x_0^{\,2} + (v_0/\omega)^2},$$ (12)

and

$$\phi =\tan^{-1}\!\left(\frac{v_0}{\omega\,x_0}\right),$$ (13)

since $\sin^2\theta+\cos^2\theta \equiv 1$ and $\tan\theta \equiv \sin\theta/\cos\theta$.

The kinetic energy of the system, which is the same as the kinetic energy of the mass, is written

$$K = \frac{1}{2}\,m\,\dot{x}^2 = \frac{1}{2}\,m\,a^2\,\omega^2\,\sin^2(\omega\,t-\phi).$$ (14)

The potential energy of the system, which is the same as the potential energy of the spring, takes the form

$$U = \frac{1}{2}\,k\,x^2= \frac{1}{2}\,k\,a^2\,\cos^2(\omega\,t-\phi).$$ (15)

Hence, the total energy is

$$E = K + U =\frac{1}{2} \,k\,a^2= \frac{1}{2}\,m\,\omega^2\,a^2,$$ (16)

since $m\,\omega^2 = k$ and $\sin^2\theta+\cos^2\theta \equiv 1$. Note that the total energy is a constant of the motion. Moreover, the energy is proportional to the amplitude squared of the oscillation. It follows, from the above expressions, that the simple harmonic oscillation of a mass on a spring is characterized by a continual back and forth flow of energy between kinetic and potential components. The kinetic energy attains its maximum value, and the potential energy its minimum value, when the displacement is zero (i.e., when $x=0$). Likewise, the potential energy attains its maximum value, and the kinetic energy its minimum value, when the displacement is maximal (i.e., when $x=\pm a$). Note that the minimum value of $K$ is zero, since the system is instantaneously at rest when the displacement is maximal.