Question

A particle is constrained to move along $x$ -axis with a velocity $u$ along the positive $x$ -axis. The acceleration ${}^{\prime }{a}^{\prime }$ of the particle varies as $a=-bx,$ where b is a positive constant and $x$ is the x-coordinate of the position of the particle. Then select the correct alternative(s)

• A. The maximum displacement of the particle from the starting point is $\frac{u}{\sqrt{b}}$
• B. The particle will oscillate about the origin
• C. Velocity is maximum at the origin
• D. Given data is insufficient to determine the exact motion of the particle.

Answer 1

Answer: (A), (B), (C)

Velocity is maximum at the origin

$\left(A\right)$

$\because \frac{dv}{dt}=-bx=v\frac{dv}{dx}$
$${\int }_u^{0}vdv={\int }_0^x-bxdx$$ (at maximum dispacement upper limit of velocity = 0)
${\Rightarrow \frac{v^2}{2}|}_u^0=-{b\frac{x^2}{2}|}_0^x$

$\Rightarrow -\frac{u^2}{2}=-\frac{b{x}^2}{2}\Rightarrow x=\frac{u}{\sqrt{b}}$

(B)

$\begin{array}{cc}F& =m(-bx)\\ a& =-bx=-{\omega }^{2}x\end{array}$
$⟹$ The particle will perform SHM

(C)
In case of SHM, velocity is maximum at the origin which is mean position of SHM.