Simple Harmonic Motion: An Introduction

Simple Harmonic Motion

We already know about oscillatory and periodic motions. Here we shall be talking about a special case of periodic motion that is simple harmonic motion. So what is simple harmonic motion? We can define simple harmonic motion as a motion in which the restoring force is directly proportional to displacement of the body from the mean position. And this restoring force always acts towards the mean position.
simple harmonic motion

Simple Harmonic Motion in a Swing

Mathematically it is represented as:
F = -kx
k is the force constant
x is the displacement from the mean position
Now we will try to find displacement as a function of time. We have,

F = -kx
ma = -kx

m\(\frac {d^2 x}{dt^2}\) + kx = 0
\(\frac {d^2 x}{dt^2}\) + \(ω^2x\) = 0

Where, ω = \( \sqrt{\frac km }\)

Solving this we will get,
x (t) = A sin (ωt+ φ)

A is the amplitude of oscillation, φ is known as the initial phase
So graphically Simple Harmonic Motion (SHM) can be represented as a sine wave as shown below.

simple harmonic motion

As at t = 0 the displacement is zero hence φ = 0° for this case. Some examples of simple harmonic motion are a pendulum, a block connected to a spring etc.

Now that we know displacement as a function of time we can find velocity and acceleration of Simple Harmonic Motion (SHM).

Velocity of SHM = \( \frac {dx}{dt}\)
\(~~~~~~~~~~~~~~~~~\) v (t) = Aω cos (ωt+ φ)

Hence velocity and displacement are 90° out of phase. Also velocity is maximum at mean position and minimum at extreme positions.

Now for acceleration,
Acceleration of SHM = \( \frac {dx}{dt}\)
\(~~~~~~~~~~~~~~~~~\) a (t) = \(-ω^2 \) x sin (ωt+ φ)

We can also say that a (t) = –\(ω^2\) x(t)

Acceleration is acting always opposite to the displacement. It is maximum at extreme positions and minimum at mean position.

For finding the time period of the Simple Harmonic Motion (SHM) we consider,

F = ma
F = -m \(ω^2 \) x (t)
Also, F = -kx (t)

Using above two equations,

m\( ω^2 \) x(t) = kx (t)
ω = \( \sqrt{\frac km} \)


T = \( \frac {2π}{ω} \)
T = \( 2π \sqrt{\frac {m}{k}}\)

So now we can see how time period is related with mass and force constant.

To learn more about SHM and its similarity with circular motion, download Byju’s the Learning App.

Before understanding similarity between SHM and circular motion, learn all about circular motion in this video.


Practise This Question

A mass is attached to a spring and it executes simple harmonic motion. The mass is also attached to a sensor such that it makes a beeping sound whenever it is in its mean position. The mass is pulled to a length d from the mean position and the frequency f of the beep is noted Now the mass is pulled to a length d' (d' > d) and the frequency of beep was noted to be f'. After this the mass was pushed to d'' (d'' < d) and the frequency of beep was noted f''. Given d, d', d'' are all close to the mean position.