Question

A force $F=(a+2x-bx)\text{N}$ is acting on a body of mass $4\text{kg},$ which is moving in a straight line. Is the body moving in simple harmonic motion ? If yes, find the equilibrium position of the particle.
[Assume $a>2,b>2$]

• A. No
• B. $\text{Yes},\frac{a}{(b+2)}$
• C. $\text{Yes},\frac{(b-2)}{a}$
• D. $\text{Yes},\frac{a}{(b-2)}$

Answer 1

The correct option is D $\text{Yes},\frac{a}{(b-2)}$

Given that,
$F=a+2x-bx$
$F=-(b-2)x+a$
$F=-(b-2)[x-\frac{a}{b-2}]$
$\Rightarrow F\propto -x$
For simple harmonic motion
$F=-kx$

So, the body is moving simple harmonically.

For equilibrium position
$F=0$
$\Rightarrow -(b-2)[x-\frac{a}{(b-2)}]=0$
$\Rightarrow x=\frac{a}{b-2}$