Equation of Simple Harmonic Motion

Equation of Simple Harmonic Motion

Simple harmonic motion as a motion in which the restoring force is directly proportional to displacement of the body from the mean position. The differential equation of simple harmonic motion is given by:

\(\frac{{{d}^{2}}x}{d{{t}^{2}}}={{\omega }^{2}}x\).

The necessary and sufficient condition for a motion to be simple harmonic is that the net restoring force must be linear, i.e. F = ma = -kx (where k is a constant). Thus,

\(a=\frac{{{d}^{2}}x}{d{{t}^{2}}}=-\frac{k}{m}x\) ; Where,

x = Instantaneous displacement.

Multiplying both the sides by dx/ dt and integrating with respect to t,

\(\left[ \frac{d}{dt}\left( \frac{dx}{dt} \right) \right]\frac{dx}{dt}=\left[ \frac{-kx}{m} \right]\frac{dx}{dt}\).

\(\int{v\frac{dv}{dt}}=\int{\frac{-k}{m}}\frac{xdx}{dt}\)                     \(\left[ \because \,\,v=\frac{dx}{dt} \right]\).


We get:

\(\frac{1}{2}{{\left(\frac{dx}{dt} \right)}^{2}}=\frac{k{{x}^{2}}}{2m}+c\); Where,

c = Constant of integration.

Now, when x is maximum dx/ dt will be zero, the maximum displacement xmax of the particle from the mean position is called amplitude and is represented by A, then the value of c comes out to be:


Here, \(\frac{1}{2}{{\left(\frac{dx}{dt} \right)}^{2}}=\frac{-k}{2m}\left( A-{{x}^{2}} \right)\),


\(\frac{k}{m}={{\omega }^{2}}\), we get:

\(\frac{dx}{dt}=\omega \sqrt{{{A}^{2}}-{{x}^{2}}}\),

This equation gives the velocity of the particle in Simple Harmonic Motion:

\(\frac{dx}{\sqrt{{{A}^{2}}-{{x}^{2}}}}=\omega dt\),

Integrating this equation with respect to t, we get:

\({{\sin }^{-1}}\frac{x}{A}=\omega t+\phi \); Where,

φ = Another constant of integration which depends on initial conditions. Thus

\(x=A\sin (\omega t+\phi)\),

Here, ω is called angular frequency and φ is called initial phase, whose value depends upon initial conditions.