Shm Position and Velocity Calculator

Physics Oscillations and Waves • Simple Harmonic Motion (shm) Basics

Written by STEM Calculators Team Published March 6, 2026 Updated March 6, 2026

Compute mass–spring simple harmonic motion: angular frequency \(\omega=\sqrt{k/m}\), period \(T\), frequency \(f\), and the motion \(x(t), v(t), a(t)\). Includes energies, an interactive plot (zoom + pan), and a spring–mass motion animation.

Spring–mass motion animation

This schematic animation uses the computed \(\omega\), \(A\), and \(\phi\). It always animates the displacement \(x(t)\).

The drawing is not to scale, but the timing matches the computed period.

Interactive plot

Mouse wheel / trackpad scroll to zoom. Drag to pan. Buttons below reset and zoom.

Animation speed 0.25× —

Tip: if your screen is narrow, scroll horizontally to see the full plot. Touch: drag to pan; pinch to zoom.

Enter values and click “Calculate”.

Rate this calculator

0.0 /5 (0 ratings)

Be the first to rate.

Your rating

Name (optional) Review (optional)

You can update your rating any time.

Theory

Simple harmonic motion (mass–spring)

A classic model for oscillations is a mass \(m\) attached to an ideal spring with spring constant \(k\). When the mass is displaced from equilibrium by a distance \(x\) and released, the spring exerts a restoring force that points back toward equilibrium. For an ideal spring (Hooke’s law), that force is \(F=-kx\). Newton’s second law then gives a second-order differential equation for the motion:

Equation of motion. Combine \(F=-kx\) with \(F=mx''\).

\[ \begin{aligned} m\,\frac{d^2x}{dt^2} &= -kx \\ \frac{d^2x}{dt^2} + \frac{k}{m}x &= 0 \end{aligned} \]

This equation says the acceleration is proportional to \(-x\), which is the defining signature of simple harmonic motion: the farther you move from equilibrium, the stronger the restoring acceleration back toward equilibrium.

Solution form, angular frequency, period

The differential equation \(x'' + (k/m)x=0\) has sinusoidal solutions. A common choice is the cosine form \[ x(t)=A\cos(\omega t+\phi), \] where \(A\ge 0\) is the amplitude and \(\phi\) is the phase constant set by initial conditions. Substituting this solution into the equation of motion forces the angular frequency \(\omega\) to satisfy \(\omega^2=k/m\). That is why a stiffer spring (larger \(k\)) produces faster oscillations, while a larger mass slows the oscillation.

Frequency relations. Connect \(\omega\), period \(T\), and frequency \(f\).

\[ \begin{aligned} \omega &= \sqrt{\frac{k}{m}} \\ T &= \frac{2\pi}{\omega} \\ f &= \frac{1}{T} \end{aligned} \]

The phase \(\phi\) shifts the oscillation in time. If you know \(x(0)\) and \(v(0)\), you can solve for \(A\) and \(\phi\) (many equivalent forms exist). In practice, it is convenient to keep \(A\) nonnegative and let \(\phi\) handle signs and timing.

Velocity, acceleration, and energy

Once \(x(t)\) is known, velocity and acceleration follow by differentiation. For the cosine solution: velocity is a sine wave shifted by \(\pi/2\), and acceleration is another cosine wave but with a factor \(-\omega^2\). This directly expresses the “restoring” character: \(a(t)=-\omega^2x(t)\).

Kinematics. Differentiate \(x(t)=A\cos(\omega t+\phi)\).

\[ \begin{aligned} v(t) &= \frac{dx}{dt} = -A\omega\sin(\omega t+\phi) \\ a(t) &= \frac{d^2x}{dt^2} = -A\omega^2\cos(\omega t+\phi) = -\omega^2 x(t) \end{aligned} \]

In an ideal, undamped oscillator, mechanical energy is conserved and shuttles between spring potential energy and kinetic energy. The potential energy stored in the spring is \(U=\tfrac12kx^2\), while kinetic energy is \(K=\tfrac12mv^2\). The total energy stays constant and equals the maximum potential energy at amplitude: \(E=\tfrac12kA^2\). When \(x\) is large, \(U\) is large and \(K\) is small; when the mass passes through equilibrium, \(x\approx 0\), potential energy is minimal and speed (hence kinetic energy) is maximal.

Energy conservation. Ideal SHM keeps \(E=U+K\) constant.

\[ \begin{aligned} U(t) &= \tfrac12 kx(t)^2 \\ K(t) &= \tfrac12 mv(t)^2 \\ E &= U+K = \tfrac12 kA^2 \quad (\text{constant}) \end{aligned} \]

These relationships also explain the common “maximum” formulas. Since \(|\sin|\le 1\) and \(|\cos|\le 1\), the maximum speed is \(v_{\max}=A\omega\) and the maximum acceleration magnitude is \(a_{\max}=A\omega^2\).

Connection to waves

Oscillations and waves are tightly linked: a wave can be viewed as many coupled oscillators distributed in space. In a transverse wave on a string, for example, each point on the string oscillates up and down (approximately like SHM) while the disturbance propagates along the string. The time-dependence at a fixed position is oscillatory, and the spatial propagation introduces wavelength \(\lambda\) and wave speed \(v\). The key kinematic relation is \(v=\lambda f\), tying the “oscillation rate” \(f\) to how fast a pattern of peaks and troughs moves. Understanding SHM (amplitude, phase, frequency, energy exchange) helps build intuition for wave behavior such as interference, resonance, and standing waves.

Frequently Asked Questions

What is angular frequency in simple harmonic motion?

Angular frequency ω measures how fast the phase advances in radians per second. For a mass–spring oscillator, ω = sqrt(k/m).

How do you find the period of a spring-mass oscillator?

First compute ω = sqrt(k/m), then use T = 2π/ω. The period depends on k and m, not on amplitude in the ideal model.

Why is acceleration proportional to negative displacement in SHM?

Hooke’s law gives a restoring force F = -kx. Using Newton’s second law m x'' = F yields x'' = -(k/m)x, so acceleration points toward equilibrium.

Is energy conserved in simple harmonic motion?

In the ideal undamped model, total energy E = (1/2)kA^2 stays constant. Energy swaps between spring potential U = (1/2)kx^2 and kinetic K = (1/2)mv^2.

Rate this calculator

Frequently Asked Questions

Related calculators