A particle moves so that its acceleration $a$ is given by $a = - bx$, where $x$ is displacement from equilibrium position and $b$ is a non-negative real constant. The time period of oscillation of the particle is

Question

Dr. Abubacker · Accepted Answer

(c) The relation between acceleration ($a$) and displacement ($x$) for a body in SHM is
$a = - \omega^{2}x$
Given, $a = - bx$
On comparing the two equations, we get
$\omega^{2} = b$
$	herefore$ $\omega = \sqrt{b}$
Since, $\omega = \frac{2\pi}{T}$
$	herefore$ $\frac{2\pi}{T} = \sqrt{b}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ T = \frac{2\pi}{\sqrt{b}}$

Q. A particle moves so that its acceleration $a$ is given by $a = - bx$, where $x$ is displacement from equilibrium position and $b$ is a non-negative real constant. The time period of oscillation of the particle is

Explanation

Reviewed by Dr. Abubacker

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