Question

Using the differential equation of linear S.H.M, derive an expression for the velocity of a particle performing linearly:

Answer 1

Velocity is the rate of change of displacement. The expression for velocity can be obtained from the expression of acceleration.

Acceleration,

$\frac{d^{2}x}{d{t}^2}=\frac{dv}{dt}=\frac{dv}{dx}\times \frac{dx}{dt}$

But, acceleration =$v\left(\frac{dv}{dt}\right)-----\left(1\right)$

But we know, $\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$----- (2)

From (1) and (2), $v\frac{dv}{dx}=-{\omega }^{2}x$

$\therefore vdv=-{\omega }^{2}x$

Integrating both side, the above equation, we get

$\int vdv=\int -{\omega }^{2}xdx=-{\omega }^2\int xdx$

Hence, $\frac{v^2}{2}=\frac{-{\omega }^2{k}^2}{2}+c$

where C is the constant of integration. Now, to find the vaue of C, lets consider boundary value condition. When a particle performing SHM is at the extreme position, displacement of the particle is maximum and velocity is zero.

$a$ is the amplitude of SHM.

Therefore, at x=$\pm a,v=0$

$\&0=\frac{-{\omega }^2{a}^2}{2}+C$

$Hence,C=\frac{{\omega }^2{a}^2}{2}$

Substituting the value of C,

$\frac{v^2}{2}=\frac{-{\omega }^2{x}^2}{2}+C$

$\therefore \frac{v^2}{2}=-\frac{{\omega }^2{x}^2}{2}+\frac{{\omega }^2{a}^2}{2}{v}^2={\omega }^2(a^2-x^2)$

Taking square root in both side, $v=\pm \omega \sqrt{a^2-x^2}$