Question

Define linear S.H.M. Obtain differential equation of linear S.H.M.

Answer 1

Linear SHM is the simplest Kind of oscillatory motion in which a body when displaced from its mean position, oscillates 'to and fro' about mean position and the restoring force (or acceleration) is always directed towards its mean position and its magnitude is directly proportional to the displacement from the mean position.

Consider particle $$P$$ moving along the circumference of a circle of radius 'a' with a uniform angular speed of $$\omega$$ in the anticlockwise direction.

Refer image .1

Particle $$P$$ along the circumference of the circle has its projection particle on diameter $$AB$$ at point $$M$$. This projection particle follows linear SHM along line $$AB$$ when particle P rotates around the circle.

Let the rotation start with initial angle $$'\alpha'$$ as shown above $$(t=0)$$

In time 't' the angle between OP and X-axis will become $$(\omega t+\alpha)$$ as shown below:

Refer image 2

Now from $$\Delta\,\,OPM=\dfrac{OM}{OP}\cos (\omega t+\alpha)$$

$$\Rightarrow \dfrac{x}{a}=\cos(\omega t+\alpha)$$

$$\Rightarrow x=a\cos (\omega t+\alpha)$$ .........(1)

Differentiating above equation with respect to time

we get velocity

$$\Rightarrow v=\dfrac{dx}{dt}=-a\omega \sin (\omega t+\alpha)$$ .........(2)

Differentiating again we get acceleration

$$\Rightarrow a=\dfrac{dv}{dt}=\dfrac{d^2x}{dt^2}-a\omega^2\cos(\omega t+\alpha)$$

$$=-a\cos(\omega t+\alpha)\omega^2$$

$$=-\omega^2x$$ ...........(3)

as $$x=a\cos (\omega t+\alpha)$$

Equation (1),(2) and (3) are differential equations for linear SHM.