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Simple harmonic motion

Consider the motion of a point particle of mass $ m$ , moving in one dimension, that is slightly displaced from a stable equilibrium point located at $ x=0$ . Suppose that the particle is moving in the conservative force field $ f(x)$ . According to the preceding analysis, in order for $ x=0$ to correspond to a stable equilibrium point, we require both

$\displaystyle f(0) = 0,$ (2.56)


$\displaystyle \frac{d f(0)}{dx} < 0.$ (2.57)

Our particle obeys Newton's second law of motion,

$\displaystyle m\,\frac{d^{\,2} x}{d t^{\,2}} = f(x).$ (2.58)

Let us assume that the particle always stays fairly close to its equilibrium point. In this case, to a good approximation, we can represent $ f(x)$ via a truncated Taylor expansion about this point. In other words,

$\displaystyle f(x) \simeq f(0) + \frac{df(0)}{dx}\,x + {\cal O}(x^{\,2}).$ (2.59)

However, according to Equations (2.56) and (2.57), the preceding expression can be written

$\displaystyle f(x) \simeq - m\,\omega_0^{\,2}\,x,$ (2.60)

where $ df(0)/dx = -m\,\omega_0^{\,2}$ . Hence, we conclude that our particle satisfies the following approximate equation of motion,

$\displaystyle \frac{d^{\,2} x}{dt^{\,2}}+ \omega_0^{\,2}\,x\simeq 0,$ (2.61)

provided that it does not stray too far from its equilibrium point; in other words, provided $ \vert x\vert$ does not become too large.

Equation (2.61) is called the simple harmonic equation; it governs the motion of all one-dimensional conservative systems that are slightly perturbed from some stable equilibrium state. The solution of Equation (2.61) is well known:

$\displaystyle x(t) = a\,\sin(\omega_0\,t -\phi_0).$ (2.62)

The pattern of motion described by this expression, which is called simple harmonic motion, is periodic in time, with repetition period $ T_0 = 2\pi/\omega_0$ , and oscillates between $ x=\pm a$ . Here, $ a$ is called the amplitude of the motion. The parameter $ \phi_0$ , known as the phase angle, simply shifts the pattern of motion backward and forward in time. Figure 2.6 shows some examples of simple harmonic motion. Here, $ \phi_0 = 0$ , $ +\pi/4$ , and $ -\pi/4$ correspond to the solid, short-dashed, and long-dashed curves, respectively.

Note that the frequency, $ \omega_0$ --and, hence, the period, $ T_0$ --of simple harmonic motion is determined by the parameters appearing in the simple harmonic equation, Equation (2.61). However, the amplitude, $ a$ , and the phase angle, $ \phi_0$ , are the two integration constants of this second-order ordinary differential equation, and are thus determined by the initial conditions; that is, the particle's initial displacement and velocity.

Figure 2.6: Simple harmonic motion.
\epsfysize =2.75in

From Equations (2.45) and (2.60), the potential energy of our particle at position $ x$ is approximately

$\displaystyle U(x) \simeq \frac{1}{2}\,m\,\omega_0^{\,2}\,x^{\,2}.$ (2.63)

Hence, the total energy is written

$\displaystyle E = K + U = \frac{1}{2}\,m\left(\frac{dx}{dt}\right)^2+ \frac{1}{2}\,m\,\omega_0^{\,2}\,x^{\,2},$ (2.64)


$\displaystyle E = \frac{1}{2}\,m\,\omega_0^{\,2}\,a^{\,2}\,\cos^2(\omega_0\,t-\...
...,a^{\,2}\,\sin^2(\omega_0\,t-\phi_0) = \frac{1}{2}\,m\,\omega_0^{\,2}\,a^{\,2},$ (2.65)

where use has been made of Equation (2.62), and the trigonometric identity $ \cos^2\theta+\sin^2\theta \equiv 1$ . Note that the total energy is constant in time, as is to be expected for a conservative system, and is proportional to the amplitude squared of the motion.

Consider the motion of a point particle of mass $ m$ that is slightly displaced from a unstable equilibrium point at $ x=0$ . The fact that the equilibrium is unstable implies that

$\displaystyle f(0) = 0,$ (2.66)


$\displaystyle \frac{d f(0)}{dx} > 0.$ (2.67)

As long as $ \vert x\vert$ remains small, our particle's equation of motion takes the approximate form

$\displaystyle m\,\frac{d^{\,2} x}{d t^{\,2}} \simeq f(0)+ \frac{df(0)}{dx}\,x,$ (2.68)

which reduces to

$\displaystyle \frac{d^{\,2}x}{dt^{\,2}} \simeq k^{\,2}\,x,$ (2.69)

where $ df(0)/dx = m\,k^{\,2}$ . The most general solution to the preceding equation is

$\displaystyle x(t) = A\,{\rm e}^{\,k\,t} + B\,{\rm e}^{-k\,t},$ (2.70)

where $ A$ and $ B$ are arbitrary constants. Thus, unless the initial conditions are such that $ A$ is exactly zero, the particle's displacement from the unstable equilibrium point grows exponentially in time.

next up previous
Next: Two-body problem Up: Newtonian mechanics Previous: Motion in one-dimensional potential
Richard Fitzpatrick 2016-03-31