Question

Equation of displacement of particle is x=2 sin mt Show that acceleration of the particle is proportional to negative of its displacement

Answer 1

The displacement of the particle is given as,

$\longrightarrow\sf{x=2\sin(\omega t)\quad\quad\dots(1)}$

where $\omega$ is the angular frequency which is constant.

We know that velocity of the particle is the first derivative of displacement wrt time.

$\longrightarrow\sf{v=\dfrac{dx}{dt}}$

$\longrightarrow\sf{v=\dfrac{d}{dt}\,[2\sin(\omega t)]}$

$\longrightarrow\sf{v=2\cdot\dfrac{d\,[\sin(\omega t)]}{d\,[\omega t]}\cdot\dfrac{d\,[\omega t]}{dt}}$

$\longrightarrow\sf{v=2\omega\cos(\omega t)}$

And we know that acceleration is the first derivative of velocity wrt time.

$\longrightarrow\sf{a=\dfrac{dv}{dt}}$

$\longrightarrow\sf{a=\dfrac{d}{dt}\,[2\omega\cos(\omega t)]}$

$\longrightarrow\sf{a=2\omega\cdot\dfrac{d\,[\cos(\omega t)]}{d\,[\omega t]}\cdot\dfrac{d\,[\omega t]}{dt}}$

$\longrightarrow\sf{a=-2\omega^2\sin(\omega t)}$

$\longrightarrow\sf{a=-\omega^2\cdot2\sin(\omega t)}$

From (1),

$\longrightarrow\sf{a=-\omega^2x}$

Since $\omega$ is a constant, so is $\omega^2.$ Hence,

$\longrightarrow\sf{\underline{\underline{a\propto -x}}}$

Hence acceleration of the particle is proportional to negative of its displacement. It also implies that retardation occurs to the particle with displacement.