Theory

Oscillations and harmonic motion

What simple harmonic motion is, how to recognize it from the equation $\ddot{x} = -\omega^2 x$, how to compute the period and energy of the spring and of pendulums, and how damping and a driving force lead to resonance.

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Periodic motion and simple harmonic motion
Equation of harmonic motion

Periodic motion and simple harmonic motion

1. Periodic motion, period and frequency

A motion is called periodic if there exists a time interval $T > 0$ such that, for every instant $t$ , the position and the velocity at the instant $t + T$ coincide with those at the instant $t$ . The smallest $T$ with this property is the period, and the number of complete cycles performed per unit time is the frequency

f = \frac{1}{T},

which is measured in hertz ( $1 Hz = 1 s^{- 1}$ ). Period and frequency are obtained by counting oscillations; for example, a pendulum that performs $30$ complete oscillations in $45 s$ has $T = 1.5 s$ and $f = 0.67 Hz$ .

2. Simple harmonic motion

A particle moving along the $x$ -axis performs simple harmonic motion if its equation of motion is

x (t) = A cos (ω t + φ),

with $A > 0$ , $ω > 0$ and $φ \in R$ constants. The number $A$ is the amplitude, that is, the maximum distance of the particle from the equilibrium position $x = 0$ ; $ω$ is the angular frequency, in $rad/s$ ; the argument $ω t + φ$ is the phase and $φ$ is the initial phase. Since the cosine has period $2 π$ , the motion is periodic with

T = \frac{2 π}{ω}, f = \frac{ω}{2 π} .

The particle oscillates between $x = - A$ and $x = A$ . A law written with the sine describes the same motion with a different initial phase, because $sin α = cos (α - \frac{π}{2})$ . For $x (t) = 0.20 cos (5 t) m$ we read $A = 0.20 m$ and $ω = 5 rad/s$ , so $T = \frac{2 π}{5} s = 1.26 s$ .

Equation of motion

x (t) = A cos (ω t)

with

A = 2

and

T = 4

: the graph lies between

- A

and

A

and repeats identically after each period

3. Velocity and acceleration in harmonic motion

In the harmonic motion $x (t) = A cos (ω t + φ)$ the velocity and the acceleration are

v (t) = \frac{d x}{d t} = - A ω sin (ω t + φ), a (t) = \frac{d v}{d t} = - A ω^{2} cos (ω t + φ) = - ω^{2} x (t) .

The formulas follow by differentiating the equation of motion twice with the chain rule. The maximum values of the magnitudes are $v_{m a x} = A ω$ , reached when passing through equilibrium, and $a_{m a x} = A ω^{2}$ , reached at the extremes $x = \pm A$ , where the velocity is zero. The relation $a = - ω^{2} x$ says that the acceleration is proportional to the displacement and opposite in direction, so it always points toward the equilibrium position. For $x (t) = 0.20 cos (5 t) m$ we have $v_{m a x} = 1.0 m/s$ and $a_{m a x} = 5.0 m/ s^{2}$ .

With

A = 1

and

T = 4

: position

x (t)

(colored), velocity

v (t)

(dashed) and acceleration

a (t)

(red); the velocity vanishes where

∣ x ∣

is maximum and the acceleration always has the opposite sign to

x

Velocity as a function of position

Time is eliminated from the laws of $x$ and $v$ . Setting $α = ω t + φ$ , we have $cos α = \frac{x}{A}$ and $sin α = - \frac{v}{A ω}$ ; since $cos^{2} α + sin^{2} α = 1$ , it follows that

\frac{x ^{2}}{A ^{2}} + \frac{v ^{2}}{A ^{2} ω ^{2}} = 1, hence ∣ v ∣ = ω A^{2} - x^{2} .

The formula gives the magnitude of the velocity at every position without knowing the instant; the direction depends on the part of the oscillation the particle is traveling. For $A = 0.10 m$ and $ω = 4 rad/s$ , at the position $x = 0.06 m$ we have $∣ v ∣ = 4 \cdot 0.08 m/s = 0.32 m/s$ .

Harmonic motion and uniform circular motion

Let $P$ be a point moving in uniform circular motion, with angular velocity $ω$ , along the circle with center $O$ and radius $A$ , and let $ω t + φ$ be the angle that $O P$ forms with the $x$ -axis at the instant $t$ . The projection $Q$ of $P$ onto the $x$ -axis has abscissa $A cos (ω t + φ)$ , so it moves in simple harmonic motion. The angular frequency of the harmonic motion is the angular velocity of the circular motion, and the phase is the angle that locates $P$ on the circle. The acceleration is also recovered, since the centripetal acceleration of $P$ has magnitude $ω^{2} A$ and points toward $O$ , and therefore its component along $x$ is $- ω^{2} A cos (ω t + φ) = - ω^{2} x$ . For uniform circular motion see the topic on motion in the plane.

The point

P

travels around the circle of radius

A

with angular velocity

ω

; its projection

Q

onto the

x

-axis has abscissa

A cos (ω t + φ)

Equation of harmonic motion

Let $ω > 0$ and denote by $\overset{x}{˙} = \frac{d x}{d t}$ and $\overset{x}{¨} = \frac{d ^{2} x}{d t ^{2}}$ the first and second derivatives with respect to time. The solutions of the differential equation

\overset{x}{¨} = - ω^{2} x

are exactly the functions $x (t) = A cos (ω t + φ)$ with $A \geq 0$ and $φ \in R$ . The theorem is used as follows. If Newton’s second law, applied to a system, leads to an equation of the form $\overset{x}{¨} = - ω^{2} x$ , then the system performs harmonic motion with angular frequency equal to the square root of the coefficient of $x$ and period $T = \frac{2 π}{ω}$ . For example, $\overset{x}{¨} = - 25 x$ describes a harmonic motion with $ω = 5 rad/s$ and $T = 1.26 s$ . The same conclusion holds for any quantity, such as an angle, that satisfies an equation of this form.

Proof

The equation can be written $\overset{x}{¨} + ω^{2} x = 0$ and is a second-order linear equation with constant coefficients. The characteristic equation $r^{2} + ω^{2} = 0$ has the complex conjugate roots $\pm iω$ , so the general solution is $x (t) = C_{1} cos (ω t) + C_{2} sin (ω t)$ with $C_{1}, C_{2} \in R$ . Set $A = C_{1}^{2} + C_{2}^{2}$ . If $A > 0$ , the point $(\frac{C _{1}}{A}, - \frac{C _{2}}{A})$ lies on the circle of radius $1$ , therefore there exists $φ$ such that $cos φ = \frac{C _{1}}{A}$ and $sin φ = - \frac{C _{2}}{A}$ . By the cosine addition formula

A cos (ω t + φ) = A cos φ cos (ω t) - A sin φ sin (ω t) = C_{1} cos (ω t) + C_{2} sin (ω t),

so every solution has the form in the statement; if $A = 0$ the solution is the zero solution. Conversely, by the theorem on velocity and acceleration in harmonic motion, every function $A cos (ω t + φ)$ has second derivative $- ω^{2} A cos (ω t + φ)$ , and therefore it satisfies the equation.

Amplitude and phase from the initial conditions

If at the instant $t = 0$ the particle is at $x_{0}$ with velocity $v_{0}$ , the amplitude and the initial phase of the harmonic motion with angular frequency $ω$ are determined by

A = x_{0}^{2} + \frac{v _{0}^{2}}{ω ^{2}}, ⎩ ⎨ ⎧ cos φ = \frac{x _{0}}{A} sin φ = - \frac{v _{0}}{A ω}

The formulas follow by imposing $x (0) = A cos φ = x_{0}$ and $v (0) = - A ω sin φ = v_{0}$ , and adding the squares of the two equalities. The two conditions on $φ$ must be satisfied together; indeed, the tangent alone, $tan φ = - \frac{v _{0}}{ω x _{0}}$ , determines $φ$ only up to $π$ , and the sign of $cos φ$ selects the correct value. A body released from rest at the position $x_{0} > 0$ has $A = x_{0}$ and $φ = 0$ ; a body launched from the equilibrium position with velocity $v_{0} > 0$ has $A = \frac{v _{0}}{ω}$ and $φ = - \frac{π}{2}$ .

The theory continues with 6 more sections (mass spring oscillator, energy of harmonic motion, simple pendulum…), plus all worked examples, the solutions and the PDF.

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