Question

Write the differential equation of simple harmonic motion

Answer 1

Answer :

During SHM,

• The force acting in the particle is directly proportional to the negative of its displacement.

• In other words,the direction of the force and displacement are mutually opposite to eachother

Now,

$\sf F \propto - x \\ \\ \longrightarrow \sf F = - kx$

But F = Ma

Thus,

$\longrightarrow \sf Ma + kx = 0$

Dividing both sides by Mass of the particle 'M' :

$\longrightarrow \sf a + \dfrac{k}{m}x = 0$

We know that,

$\sf a = \dfrac{dx^2}{d^2t} \ and \ \omega^2 = \dfrac{k}{m}$

Thus,

$\longrightarrow \sf \dfrac{dx^2}{d^2t} + \omega^2 x = 0$

The general equation of SHM is a consequence of solving the above differential equation

Answer 2

➡ Derivation of differential equation of Simple Harmonic Motion (SHM) :

✏ First of all, In SHM force acted on particle is directly proportional to displacement from mean position.

✏ It is free-fro type of motion about mean position.

✏ Direction of force is opposite to the direction of displacement.

$\therefore \rm \: \red{ F \propto - X}$

✏ putting k as constant in equation...

$\implies \rm \: \blue{F = - kX} \: \longrightarrow \: (1)$

✏ Net force on particle is given by...

$\implies \rm \: \pink{F = ma} \longrightarrow(2)$

✏ By adding both equations...

$\implies \rm \: ma +kX= 0$

✏ Divide this equation with mass of particle...

$\implies \rm \: a + \frac{k}{m} X = 0 \\ \\ \dag \rm \: we \: know \: that \: \green{a = \frac{ {d}^{2}X }{d {t}^{2} } } \: and \: \purple{ { \omega}^{2} = \frac{k}{m} }\\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{ \frac{ {d}^{2}X }{d {t}^{2} } + { \omega}^{2} X = 0}}}}}$