harmonic oscillator kinematics and dynamics

A harmonic oscillator is any system whose restoring force is proportional to the displacement from equilibrium and directed back toward it: $F = -kx$.

Newton's second law then gives $ma = -kx$, so the acceleration obeys $a = -\omega^2 x$, where $\omega = \sqrt{\frac{k}{m}}, \qquad \omega = \frac{2\pi}{T};$ this relation is the dynamics of the oscillator. Solving that equation gives the kinematics of the motion: $x(t) = A\cos(\omega t + \varphi).$ Differentiating twice yields $v(t) = -A\omega\sin(\omega t + \varphi), \qquad a(t) = -A\omega^2\cos(\omega t + \varphi) = -\omega^2 x(t).$ The amplitude $A$ and the initial phase $\varphi$ are fixed by the initial position and velocity, which relies on the prerequisites concerning potential energy and conservation of mechanical energy. Because the force is conservative with $U = \tfrac12 kx^2$, the total mechanical energy $E = \tfrac12kx^2 + \tfrac12mv^2 = \tfrac12kA^2 = \text{const}$ is conserved, which lets you find speeds without solving the motion in time.

A mass on a spring oscillates as $x(t) = A\cos(\omega t)$ with $A = 0.10\ \text{m}$ and period $T = 2.0\ \text{s}$. Find $x$, $v$ and $a$ at $t = 0.50\ \text{s}$.

A block of mass $m = 0.50\ \text{kg}$ is attached to a spring of stiffness $k = 8.0\ \text{N/m}$ and released from rest at $x_0 = 0.040\ \text{m}$. Find $\omega$, the period $T$, the maximum speed, and the speed at $x = 0.020\ \text{m}$.

The same oscillator ($m = 0.50\ \text{kg}$, $k = 8.0\ \text{N/m}$, hence $\omega = 4.0\ \text{rad/s}$) is started with $x_0 = 0.030\ \text{m}$ and $v_0 = -0.40\ \text{m/s}$. Write $x(t)$ and find the first time after $t = 0$ at which the block passes through equilibrium.

Use the PhET Masses and Springs simulation or an equivalent mass–spring model; do the observing and measuring in the simulation rather than with a real hanging mass.

1. Set $k = 8.0\ \text{N/m}$ and $m = 0.50\ \text{kg}$. Before running, predict $T = 2\pi\sqrt{m/k}$ and write the prediction down.
2. Pull the mass to $A = 0.040\ \text{m}$, release it, and record the $x(t)$ graph. Measure the period from successive crossings of equilibrium and compare it with your prediction.
3. Repeat with $A = 0.020\ \text{m}$ and $A = 0.060\ \text{m}$. Note whether the period changes with amplitude.
4. Record the speed at $x = 0$ and at $x = A$; compare with $A\omega$ and with zero.
5. Plot $x$, $v$ and $a$ over the same time interval and check the phase shifts: $v$ leads $x$ by $T/4$, and $a$ is opposite to $x$.
6. Change the mass to $2.0\ \text{kg}$; predict the new period first, then measure it and compare with $2\pi\sqrt{m/k}$.

Show how the position, velocity and acceleration of a harmonic oscillator are related in time: $v$ leads $x$ by a quarter period, and $a$ is opposite in sign to $x$ and obeys $a = -\omega^2 x$.

Three vertically stacked graphs sharing one horizontal time axis for one full period $T$: $x(t) = A\cos(\omega t)$, $v(t) = -A\omega\sin(\omega t)$ and $a(t) = -A\omega^2\cos(\omega t)$. A dashed vertical line at $t = T/8$ connects the values on all three graphs; beside them is a small sketch of a mass on a spring with the $x$-axis and the equilibrium position marked.

The front axle of a $1400\ \text{kg}$ car carries $700\ \text{kg}$, so each of its two springs supports $m = 350\ \text{kg}$ with stiffness $k = 2.8\times10^{4}\ \text{N/m}$. A speed bump pushes the body down by $A = 5.0\ \text{cm}$ and lets go. The angular frequency is $\omega=\sqrt{k/m}=8.9\ \text{rad/s}$, so the body bounces with period $T=2\pi\sqrt{m/k}=0.70\ \text{s}$, while the largest speed and acceleration reach $v_{\max}=\omega A=0.45\ \text{m/s}$ and $a_{\max}=\omega^{2}A=4.0\ \text{m/s}^{2}$ (about $0.4g$).

This is a textbook harmonic oscillator: the restoring force is linear, $F=-kx$, and Newton's second law gives $a=-\frac{k}{m}x$, whose solution is $x=A\cos(\omega t)$ with $\omega=\sqrt{k/m}$. The scenario exercises the kinematic pair $v=\dot x$ and $a=\ddot x$, which lag a quarter period behind the displacement, and the fact that $T$ does not depend on the $5.0\ \text{cm}$ amplitude.

The A2 string of a guitar sounds at $f = 110\ \text{Hz}$. Its vibrating length is $0.65\ \text{m}$ with linear density $1.5\ \text{g/m}$, giving $m = 9.8\times10^{-4}\ \text{kg}$; plucked at the middle it swings with amplitude $A = 1.0\ \text{mm}$. Then $\omega = 2\pi f = 691\ \text{rad/s}$, the effective stiffness is $k = m\omega^{2} = 4.7\times10^{2}\ \text{N/m}$, the middle of the string moves at up to $v_{\max}=\omega A = 0.69\ \text{m/s}$, and its acceleration reaches $a_{\max}=\omega^{2}A = 4.8\times10^{2}\ \text{m/s}^{2}\approx 49g$. The stored mechanical energy is $E=\tfrac12 kA^{2}=2.3\times10^{-4}\ \text{J}$.

The plucked string obeys the same kinematics as a mass on a spring: $x = A\sin(\omega t)$, $v = \omega A\cos(\omega t)$, $a = -\omega^{2}x$. The numbers make the phase relations visible — the speed peaks at $x=0$, while the acceleration peaks at the extreme deflection of $1.0\ \text{mm}$, where the speed is zero.

A phone in a jogger's pocket records the vertical motion of the torso: the body rises and falls twice per stride, $f = 2.0\ \text{Hz}$, with amplitude $A = 3.0\ \text{cm}$. The built-in accelerometer samples at $100\ \text{Hz}$ and sees a sinusoidal signal $a(t)$: $\omega = 2\pi f = 12.6\ \text{rad/s}$ and a peak value $a_{\max}=\omega^{2}A = 4.7\ \text{m/s}^{2}\approx 0.48g$. The torso's speed peaks at $v_{\max}=\omega A = 0.38\ \text{m/s}$.

The torso is treated as a harmonic oscillator, so its acceleration $a=-\omega^{2}x$ is proportional to the displacement and always points back towards the middle of the step — that is why the accelerometer trace is sinusoidal and why its amplitude grows as the square of the step frequency. Counting zero crossings of that trace gives the step rate, because one full cycle of $x=A\cos(\omega t)$ corresponds to one stride.