Question

For the given oscillatory motion, $F=-k(x-b{)}^3$, find out the mean position of the particle.

• A. $x=-b$
• B. $x=+b$
• C. $x=0$
• D. $x=\frac{b}{2}$

Answer 1

The correct option is B $x=+b$

The mean position is the point of stable equilibrium.
$\Rightarrow F_{net}=0$ at mean position
$\Rightarrow -k(x-b{)}^3=0$
$\Rightarrow (x-b{)}^3=0$
$\Rightarrow x=b$