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Formula

Show calculation steps (3)
  1. Velocity v(t)

    Velocity v(t): Simple Harmonic Motion Calculator

    v = -A ω sin(ω t + φ), with ω = 2π f

  2. Acceleration a(t)

    Acceleration a(t): Simple Harmonic Motion Calculator

    a = -A ω² cos(ω t + φ), with ω = 2π f

  3. Period T

    Period T: Simple Harmonic Motion Calculator

    T = 1 / f

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Results

Displacement x(t)
0.1545
meters
Velocity v(t) -5.9757 m/s
Acceleration a(t) -24.399 m/s²
Angular frequency ω 12.5664 rad/s
Period T 0.5 s

What is Simple Harmonic Motion?

Simple harmonic motion (SHM) describes any oscillation in which the restoring force is proportional to displacement, such as a mass on a spring or a small-angle pendulum. Its position over time follows a cosine wave. This calculator returns the displacement, velocity, acceleration, angular frequency, and period at any instant from four inputs: amplitude A, frequency f, phase φ, and time t.

Mass on a horizontal spring oscillating between two extremes around an equilibrium point
A mass on a spring oscillating about its equilibrium position is a classic example of simple harmonic motion.

How to use it

Enter the amplitude in meters, the frequency in hertz, the phase angle in radians, and the time in seconds. The calculator computes \( \omega = 2\pi f \), then evaluates the displacement, velocity and acceleration equations at your chosen instant. All results use SI units.

The formula explained

The core equation is $$x(t) = A\cos\!\left(\omega t + \varphi\right)$$ where \( \omega = 2\pi f \) is the angular frequency in rad/s. Differentiating once gives velocity $$v(t) = -A\omega\sin\!\left(\omega t + \varphi\right)$$ and again gives acceleration $$a(t) = -A\omega^{2}\cos\!\left(\omega t + \varphi\right) = -\omega^{2} x$$ The period \( T = \dfrac{1}{f} \) is the time for one complete cycle.

Displacement, velocity and acceleration sinusoidal curves of simple harmonic motion over time
Displacement follows a cosine curve; velocity and acceleration are phase-shifted sinusoids.

Worked example

Suppose \( A = 0.5 \) m, \( f = 2 \) Hz, \( \varphi = 0 \), and \( t = 0.1 \) s. Then $$\omega = 2\pi(2) \approx 12.566 \ \text{rad/s}$$ and the argument is \( \omega t = 1.2566 \) rad. Displacement $$x = 0.5\cdot\cos(1.2566) \approx 0.1545 \ \text{m}$$ Velocity $$v = -0.5\cdot 12.566\cdot\sin(1.2566) \approx -5.975 \ \text{m/s}$$ Acceleration $$a = -0.5\cdot 12.566^{2}\cdot\cos(1.2566) \approx -24.40 \ \text{m/s}^{2}$$ The period \( T = \dfrac{1}{2} = 0.5 \) s.

FAQ

Why does phase use radians? The cosine argument is an angle, so \( \varphi \) and \( \omega t \) must share the same unit; radians are standard in physics.

What if frequency is zero? A zero frequency means no oscillation, so \( \omega = 0 \) and the period is undefined (shown as 0).

Can I use it for a pendulum? Yes, for small angles a pendulum is approximately SHM; use its natural frequency for \( f \).

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