# Simple Harmonic Motion
Source: algebrica.org - CC BY-NC 4.0
https://algebrica.org/simple-harmonic-motion/
Fetched from algebrica.org post 14882; source modified 2026-03-20T22:29:00.
## What is simple harmonic motion
A simple harmonic motion is a straight-line motion obtained by projecting the uniform circular motion of a body onto a fixed diameter of the circle. In this case, a material point repeatedly moves along the diameter, oscillating back and forth, and returning to the same position at regular [intervals](../intervals.md) of time equal to the period of the motion.
This projection creates a smooth and continuous oscillation, where the displacement from the center varies [sinusoidally](../sine-and-cosine.md) with time. The position as a function of time for simple harmonic motion is given by the equation:
\\[
x(t) = A \\sin(\\omega t)
\\]
- \\( A \\) is the amplitude, the maximum displacement from the equilibrium position (the function \\( \\sin(\\omega t) \\) oscillates between the values \\( +1 \\) and \\( -1 \\)).
- \\( \\omega \\) is the angular frequency (in radians per second).
- \\( t \\) is the time variable.
Since the function is periodic with period \\( T \\), meaning it repeats itself from \\( t \\) to \\( t+T \\), the argument of the trigonometric function must change by \\( 2\\pi \\).
Therefore, we have:
\\[
\\omega(t+T) - \\omega t = 2\\pi
\\]
From this, we derive the fundamental relation:
\\[
\\omega = \\frac{2\\pi}{T}
\\]
where \\( \\nu \\) represents the frequency of the harmonic motion, and is related to the period by:
\\[
\\nu = \\frac{1}{T}
\\]
* * *
In simple harmonic motion where the center of oscillation does not coincide with the origin but is located at \\( x\_0 \\), the general equation of motion is:
\\[
x - x\_0 = A \\sin(\\omega t + \\varphi)
\\]
- \\( x(t) \\) is the position of the material point at time ( t ),
- \\( x\_0 \\) is the center around which the motion oscillates,
- \\( A \\) is the amplitude,
- \\( \\omega \\) is the angular frequency,
- \\( \\varphi \\) is the initial phase.
## Example 1
For example, calculate the equation of a simple harmonic motion with period \\( T = 3 \\, \\text{s} \\) and amplitude \\( A = 0.5 \\, \\text{m} \\).
We find the angular frequency.
\\[
\\omega = \\frac{2\\pi}{T} = \\frac{2\\pi}{3} \\, \\text{rad/s}
\\]
Thus, the motion equation is:
\\[
x(t) = 0.5 \\sin\\left( \\frac{2\\pi}{3} t \\right)
\\]
## Velocity
To find the instantaneous [scalar velocity](../velocity.md) \\( v(t) \\), we [differentiate](../derivatives.md) \\( x(t) \\) with respect to time:
\\[
v(t) = \\frac{dx}{dt}
\\]
Applying the derivative:
\\[
\\frac{d}{dt} \\left( A \\sin(\\omega t) \\right) = A \\omega \\cos(\\omega t)
\\]
Thus, the instantaneous velocity of the material point is:
\\[
v(t) = A \\omega \\cos(\\omega t)
\\]
This expression shows that the velocity varies over time following a [cosine function](../cosine-function.md), reaching its maximum when the particle passes through the equilibrium position.
- The velocity in simple harmonic motion is not constant: it changes over time following the trend of a cosine function.
- When the body passes through the equilibrium position (that is, \\(x = 0\\), the center of motion), the velocity is maximum.
- When the body reaches the extremes (that is, at the points \\(x = +A\\) or \\(x = -A\\)), the velocity is zero.
- The velocity, being a function of the cosine, is phase-advanced with respect to the displacement \\(x\\) in simple harmonic motion.
## Example 2
For example, calculate the instantaneous velocity of a simple harmonic motion with period \\( T = 3 \\, \\text{s} \\) and amplitude \\( A = 0.5 \\, \\text{m} \\).
* * *
We find the angular frequency.
\\[
\\omega = \\frac{2\\pi}{T} = \\frac{2\\pi}{3} \\, \\text{rad/s}
\\]
The position function is:
\\[
x(t) = 0.5 \\sin\\left( \\frac{2\\pi}{3} t \\right)
\\]
Differentiating with respect to time to find the velocity:
\\[
v(t) = \\frac{dx}{dt} = 0.5 \\times \\frac{2\\pi}{3} \\cos\\left( \\frac{2\\pi}{3} t \\right)
\\]
Simplifying we obtain:
\\[
v(t) = \\frac{\\pi}{3} \\cos\\left( \\frac{2\\pi}{3} t \\right)
\\]
## Acceleration
To find the instantaneous [acceleration](../acceleration.md) \\( a(t) \\), we [differentiate](../derivatives.md) the velocity function \\( v(t) \\) with respect to time:
\\[
a(t) = \\frac{dv}{dt}
\\]
Applying the derivative:
\\[
\\frac{d}{dt} \\left( A \\omega \\cos(\\omega t) \\right) = -A \\omega^2 \\sin(\\omega t)
\\]
Thus, the instantaneous acceleration of the material point is:
\\[
a(t) = -A \\omega^2 \\sin(\\omega t)
\\]
This expression shows that the acceleration varies over time following a [sine function](../sine-function.md), reaching its maximum magnitude when the particle is at the maximum displacement from the equilibrium position.
- The acceleration in simple harmonic motion is not constant: it changes over time following the trend of a sine function.
- When the body passes through the equilibrium position (that is, \\(x = 0\\), the center of motion), the acceleration is zero.
- When the body reaches the extremes (that is, at the points \\(x = +A\\) or \\(x = -A\\)), the acceleration reaches its maximum magnitude.
The acceleration, being a function of the sine, is in phase opposition with respect to the displacement \\(x\\). In simple harmonic motion the acceleration has only a normal component \\( a\_n \\) and it is centripetal, always directed toward the center of oscillation. This means that as the particle moves, the acceleration acts continuously to pull it back toward the equilibrium position.
> In uniformly [accelerated rectilinear motion](../acceleration.md), the acceleration [vector](../vectors.md) has two components: a normal component \\( a\_n \\) and a tangential component \\( a\_t \\). Since the trajectory is still a straight line, the normal component \\( a\_n = 0 \\). However, the tangential component \\( a\_t \\) is constant and nonzero, representing a steady change in the velocity’s magnitude along the direction of motion.
## Example 3
For example, calculate the acceleration in a simple harmonic motion with period \\( T = 3 , \\text{s} \\) and amplitude \\( A = 0.5 \\, \\text{m} \\).
* * *
Let’s review the steps followed in the previous examples. First, we find the angular frequency.
\\[
\\omega = \\frac{2\\pi}{T} = \\frac{2\\pi}{3} \\, \\text{rad/s}
\\]
Starting from the motion equation:
\\[
x(t) = 0.5 \\sin\\left( \\frac{2\\pi}{3} t \\right)
\\]
The velocity is:
\\[
v(t) = 0.5 \\times \\frac{2\\pi}{3} \\cos\\left( \\frac{2\\pi}{3} t \\right) = \\frac{\\pi}{3} \\cos\\left( \\frac{2\\pi}{3} t \\right)
\\]
Differentiating again, we find the acceleration:
\\[
a(t) = \\frac{dv}{dt} = -\\left( \\frac{\\pi}{3} \\times \\frac{2\\pi}{3} \\right) \\sin\\left( \\frac{2\\pi}{3} t \\right)
\\]
Simplifying:
\\[
a(t) = -\\frac{2\\pi^2}{9} \\sin\\left( \\frac{2\\pi}{3} t \\right)
\\]
