Period frequency amplitude and phase

A harmonic oscillation repeats in time. Its period $T$ is the duration of one complete cycle, and its frequency $f$ is the number of cycles per second, so $T=\frac{1}{f}$ and $\omega=2\pi f=\frac{2\pi}{T}$.
The amplitude $A$ is the maximum displacement from equilibrium and sets the scale of the motion; for a mass–spring oscillator the displacement can be written $x(t)=A\cos(\omega t+\varphi)$.
The phase $\omega t+\varphi$ fixes the state of the oscillator at a given instant, while the initial phase $\varphi$ selects where in the cycle the motion starts at $t=0$.
Because the restoring force is conservative, the total mechanical energy $E=K+U$ is constant, and for an ideal spring $E=\frac{1}{2}kA^2$; kinetic and potential energy exchange twice per period.
Thus changing $A$ changes the stored energy, while $T$ and $f$ for an ideal mass–spring system are set by $m$ and $k$ through $\omega=\sqrt{\frac{k}{m}}$.

The oscillation is given by $x(t)=0.10\cos\!\left(4\pi t+\frac{\pi}{3}\right)$ (in meters). Find the amplitude $A$, angular frequency $\omega$, frequency $f$, period $T$, initial phase $\varphi$, and the displacement at $t=0$.

A mass $m=0.50$ kg is attached to a spring of stiffness $k=8.0$ N/m. The oscillation amplitude is $A=0.12$ m. Find the period $T$, frequency $f$, total mechanical energy $E$, and maximum speed $v_{\max}$.

A mass–spring oscillator has $m=0.20$ kg and $k=20$ N/m. The total mechanical energy of oscillation is $E=0.10$ J. At $t=0$ the displacement is $x=0.05$ m, and the mass is moving toward equilibrium. Find the amplitude $A$, angular frequency $\omega$, period $T$, frequency $f$, initial phase $\varphi$, and write the equation $x(t)$.

Use a safe simulation such as PhET Masses and Springs or Pendulum Lab.

1. Set the spring constant to $k=10$ N/m and the mass to $m=0.50$ kg.
2. For amplitudes $A=0.02$ m, $0.04$ m, and $0.08$ m, measure the period $T$ and record whether $T$ changes with $A$.
3. Compute $f=1/T$ and compare with $\omega/(2\pi)$ using $\omega=\sqrt{k/m}$.
4. Add a second identical oscillator and start one at maximum displacement and the other at zero displacement moving in the positive direction. Sketch both displacement–time graphs and determine their phase difference.
   There is no hazard; use the simulation rather than dangerous setups.

To show how amplitude, period, frequency, and phase are encoded in a displacement–time graph and how a phase shift appears as a horizontal shift between two curves.

A displacement–time graph with two sinusoidal curves $x_1(t)=A\cos(\omega t)$ and $x_2(t)=A\cos(\omega t+\varphi)$. Mark $A$ as the vertical maximum, $T$ as the time between successive peaks, and show the phase shift $\varphi$ as a horizontal shift. Include a simple sketch of a mass–spring oscillator at maximum displacement to indicate amplitude.

A gymnast on a trampoline repeats one full up-down cycle every $0.60\ \mathrm{s}$. The centre of mass moves from $0.35\ \mathrm{m}$ above equilibrium to $0.35\ \mathrm{m}$ below it, so the amplitude is $A=0.35\ \mathrm{m}$. The frequency is $f=1/T=1/(0.60\ \mathrm{s})=1.67\ \mathrm{Hz}$. A second gymnast bounces with the same amplitude but is half a cycle behind: the phase difference is $\Delta\phi=\pi\ \mathrm{rad}$; when the first is at the top, the second is at the bottom.

This uses the period-frequency relation $T=1/f$, amplitude as maximum displacement from equilibrium, and phase difference: a delay of half a period gives $\Delta\phi=\pi\ \mathrm{rad}$.

A smartphone vibration motor spins at $12000\ \mathrm{rpm}=200\ \mathrm{Hz}$, so its period is $T=1/(200\ \mathrm{Hz})=5.0\ \mathrm{ms}$. The tiny moving mass has displacement amplitude $A=0.03\ \mathrm{mm}$. A headphone plays a $440\ \mathrm{Hz}$ note; the period is $T=1/(440\ \mathrm{Hz})=2.27\ \mathrm{ms}$ and the cone amplitude is $A=0.10\ \mathrm{mm}$. If the left signal is delayed by $1.14\ \mathrm{ms}$ (about half a period), the phase difference is $\Delta\phi=2\pi\Delta t/T\approx\pi\ \mathrm{rad}$, and at that frequency the sound at the ear is reduced.

It uses frequency as cycles per second, period $T=1/f$, amplitude as maximum displacement, and phase shift $\Delta\phi=2\pi\Delta t/T$ to compare two oscillations.

A car of mass $1200\ \mathrm{kg}$ crosses a speed bump. The body oscillates at $f=1.25\ \mathrm{Hz}$, so $T=1/f=0.80\ \mathrm{s}$. The initial vertical amplitude is $A_0=0.08\ \mathrm{m}$, and after two cycles it falls to $A=0.02\ \mathrm{m}$ because of the dampers. The front wheels meet the bump $0.25\ \mathrm{s}$ before the rear wheels; with $T=0.80\ \mathrm{s}$, the rear body motion lags in phase by $\Delta\phi=2\pi(0.25/0.80)=1.96\ \mathrm{rad}\approx112^\circ$.

Here amplitude sets the maximum displacement and the mechanical energy scale, period and frequency set the timing of the oscillation, and phase gives the lag between the front and rear body motions.