## Simple Harmonic Motion

Mathematically it is represented as:

F = -kx

Where,

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

\(\frac {d^2 x}{dt^2}\)

Where, ω = \( \sqrt{\frac km }\)

Solving this we will get,

x (t) = A sin (ωt+ φ)

So graphically Simple Harmonic Motion (SHM) can be represented as a sine wave as shown below.

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 = \( \frac {dx}{dt}\)

\(~~~~~~~~~~~~~~~~~\)

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

Acceleration of SHM = \( \frac {dx}{dt}\)

\(~~~~~~~~~~~~~~~~~\)

We can also say that a (t) = –\(ω^2\)

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 \(ω^2 \)

Also, F = -kx (t)

Using above two equations,

m\( ω^2 \)

ω = \( \sqrt{\frac km} \)

Therefore,

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.

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

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