Physics - SHM Question with Solution | TestHub

$U=-4(x-1{)}^2(x-2{)}^2$ ..(i)

$\frac{dU}{dx}=-4\left[2(x-1)(x-2{)}^2+2(x-1{)}^2(x-2)\right]$

Or, $\frac{dU}{dx}=-8(x-1)(x-2)[x-2+x-1]=-8(x-1)(x-2)(2x-3)$

Force on the particle is $F=-\frac{dU}{dx}=8(x-1)(x-2)(2x-3)$ ..(ii)

Force (i.e., $\frac{dU}{dx}$ ) is zero at $x=1,2,1.5\mathrm{m}$. These are positions of equilibrium. Nature of the equilibrium can be ascertained by finding $\frac{d^{2}U}{d{x}^2}$ or just by plotting the graph of $U$ versus $x$.

A) From (ii), the force on the particle is negative at $x=-1m$. If it is released at

$x=-1m$, it will move in the negative $x$ direction.

B) If the particle is released at $x=1m$, it will remain in unstable equilibrium.

C) Force $F=8(x-1)(x-2)(2x-3)$

When the particle is displaced $\Delta x$ from its equilibrium position at $x=1.5\mathrm{m}$ (so that its co-ordinate is $1.5+\Delta x)$, the force on it is

$F=8(1.5+\Delta x-1)(1.5+\Delta x-2)(3+2\Delta x-3)$

$=8(0.5+\Delta x)(\Delta x-0.5)(2\Delta x)$

$=8\left(\Delta x^2-0.5^2\right)(2\Delta x)=16\Delta x\left(\Delta x^2-0.5^2\right)$

If $\Delta x\ll 0.5$, then $\Delta x^2-0.5^2\approx -0.5^2$ Or,

$F=16\Delta x\left(-0.5^2\right)=-4\Delta x$

$\therefore ma=-4\Delta x\Rightarrow a=-4\Delta x[\because m=1kg]$

∴ Motion is SHM and $\omega =\sqrt{4}=2$

D) It is same as (C)