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7.1 Defining Simple Harmonic Motion (SHM)

Identify the defining relationship

Simple harmonic motion (SHM) is periodic motion produced by a restoring force whose magnitude is proportional to displacement from equilibrium and whose direction is opposite that displacement.

The equilibrium position is where the net force is zero. Measure signed displacement $\Delta x$ from this position; the force always points back toward it.

Map displacement to restoring direction

Position relative to equilibrium	$\Delta x$	Restoring force $F_x$
Right/positive side	Positive	Negative, toward equilibrium
At equilibrium	Zero	Zero
Left/negative side	Negative	Positive, toward equilibrium

Use the linear restoring model

$ma_x=-k\Delta x$

Calculate force and acceleration

Example: a $0.50\,\text{kg}$ object is displaced $+0.12\,\text{m}$ in a system with $k=25\,\text{N}\,\text{m}^{-1}$ .

$F_x=-k\Delta x=-(25)(0.12)=-3.0\,\text{N}$

$a_x=F_x/m=-3.0/0.50=-6.0\,\text{m}\,\text{s}^{-2}$ .

Both negative signs mean the force and acceleration point back toward equilibrium.

Separate SHM from other periodic motion

Periodic motion alone is not enough: SHM requires the linear, opposite restoring relationship. A pendulum can be modeled as SHM only for small angular displacements, when restoring torque is proportional to angular displacement.

7.2 Frequency and Period of SHM

Connect cycle time and cycle rate

The period $T$ is the time for one complete cycle, measured in seconds. The frequency $f$ is the number of cycles per second, measured in hertz.

$\begin{gathered}T=\frac1f\\ f=\frac1T\end{gathered}$

Choose the oscillator model

SHM model	Period	Parameters that set the period
Object–ideal-spring oscillator	$T_s=2\pi\sqrt{m/k}$	Oscillating mass $m$ ; spring constant $k$
Small-angle simple pendulum	$T_p=2\pi\sqrt{\ell/g}$	Pendulum length $\ell$ ; gravitational field strength $g$

Calculate spring period and frequency

Spring example: $m=0.50\,\text{kg}$ and $k=200\,\text{N}\,\text{m}^{-1}$ .

$T_s=2\pi\sqrt{\dfrac{0.50}{200}}=2\pi(0.050)=0.314\,\text{s}$

$f=\dfrac1T=\dfrac1{0.314}=3.18\,\text{Hz}$ .

The unit check works because $m/k$ has units of $\text{s}^2$ .

Predict square-root changes

Parameter change	Period change
Spring mass $m$ doubled	$T_s$ multiplied by $\sqrt2$
Spring constant $k$ multiplied by $4$	$T_s$ divided by $2$
Pendulum length $\ell$ multiplied by $4$	$T_p$ multiplied by $2$
Field strength $g$ multiplied by $4$	$T_p$ divided by $2$

Respect each model's conditions

Match the equation to the physical oscillator. The pendulum result assumes a small angular displacement; the spring result assumes an ideal spring. Frequency changes inversely with period, so a longer period means a lower frequency.

7.3 Representing and Analyzing SHM

Represent displacement sinusoidally

$x=A\cos(2\pi ft)\quad\text{or}\quad x=A\sin(2\pi ft)$

$A$ is maximum displacement from equilibrium and $f$ sets the cycle rate. Sine and cosine describe the same SHM with different choices of where the clock starts.

Match the three key positions

Position	Velocity	Acceleration
$x=+A$	$v=0$	Maximum magnitude toward negative/equilibrium direction
$x=0$	Maximum speed	$a=0$
$x=-A$	$v=0$	Maximum magnitude toward positive/equilibrium direction

Trace one cosine-start cycle

Time for cosine start	Displacement	Motion state
$0$	$+A$	Released from rest; acceleration toward equilibrium
$T/4$	$0$	Moving fastest in the negative direction; acceleration zero
$T/2$	$-A$	Instantaneously at rest; acceleration toward equilibrium
$3T/4$	$0$	Moving fastest in the positive direction; acceleration zero
$T$	$+A$	One complete cycle

Read zeros and extrema

On an $x$ –time graph, velocity is zero at displacement peaks and greatest in magnitude where the curve crosses equilibrium. Acceleration always has the opposite sign to $x$ , so its extrema align with displacement extrema but point oppositely.

Change amplitude without changing period

Increasing amplitude makes the displacement graph taller, but it does not change the period of an ideal SHM system. Do not infer a longer cycle merely because the object travels farther.

7.4 Energy of Simple Harmonic Oscillators

Hold total energy constant

$E_{\mathrm{total}}=U+K=\text{constant}$

Match energy to position

Position	Speed	Kinetic energy $K$	Potential energy $U$
Equilibrium, $x=0$	Maximum	Maximum	Minimum
Between equilibrium and a turning point	Intermediate	Between min and max	Between min and max
Turning point, $x=\pm A$	Zero	Minimum: $0$	Maximum: $E_{\mathrm{total}}$

Use the spring energy model

$\begin{gathered}E_{\mathrm{total}}=\frac12kA^2\\ U(x)=\frac12kx^2\\ K(x)=E_{\mathrm{total}}-U(x)\end{gathered}$

Calculate the energy split

Spring example: $k=80\,\text{N}\,\text{m}^{-1}$ and amplitude $A=0.20\,\text{m}$ . At $x=0.12\,\text{m}$ :

$E_{\mathrm{total}}=\tfrac12(80)(0.20)^2=1.60\,\text{J}$

$U=\tfrac12(80)(0.12)^2=0.576\,\text{J}$

$K=1.60-0.576=1.02\,\text{J}$ to three significant figures. The two contributions still total $1.60\,\text{J}$ .

Predict amplitude effects

Amplitude change	Spring total-energy change
$A\to2A$	$E_{\mathrm{total}}\to4E_{\mathrm{total}}$
$A\to A/2$	$E_{\mathrm{total}}\to E_{\mathrm{total}}/4$

Do not confuse one form with the total

At equilibrium, potential energy is minimum but the system's total energy is not zero: kinetic energy is maximum. At a turning point, kinetic energy is zero but total energy remains stored as potential energy.

7 Oscillations

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