Question

Derive the expressions for displacement, velocity and acceleration of a particle executes S.H.M.

Answer 1

Let's consider a particle P moves on the circumference of circle of radius A with uniform acceleration $\omega$.Let PN is the perpendicular drawn from P to diameter yy'.

see figure, $\angle{POX}=\theta$, OP = A and ON = y
from triangle ∆ONP,
$sin\theta=\frac{ON}{OP}$
we know, $\theta=\omega t$

so, $OPsin\omega t=ON$

or, $\boxed{y=Asin\omega t}$ ....(i),This is the expression of equation of SHM.
now differentiate y with respect to t, to find velocity of particle.
$\frac{dy}{dt}=v=\omega Acos\omega t$
or, $\boxed{v=\omega Acos\omega t}$ .....(ii)
from equations (i) and (ii),
$\boxed{\bf{v=\omega\sqrt{A^2-y^2}}}$ this is the expression of velocity of particle in SHM.

again , differentiate v with respect to t ,
$\frac{dv}{dt}=a=-\omega^2Asin\omega t$
from equation (I),
$\boxed{\bf{a=-\omega^2y}}$ this is the expression of acceleration of particle in SHM.