The potential energy of a particle is given by formula $U=100-5x+100x^2$, where U and x are in SI units. If mass of the particle is 0.1 kg then magnitude of it’s acceleration

At mean position force is zero.

$\mathrm{F}=-\frac{\mathrm{dU}}{\mathrm{dx}}=5-200\mathrm{x}$

$\Rightarrow \mathrm{x}=0.025 \mathrm{m}\text{ is mean position }$

$\text{ Acceleration, }\mathrm{a}=\frac{\mathrm{F}}{\mathrm{m}}=10(5-200\mathrm{x})$

Case 1 : At 0.05 m from origin

If x=+0.05m,

$\mathrm{a}=-50 \mathrm{m}/{\mathrm{s}}^2$

Negative sign implies direction is towards mean position.

If x=-0.05m,

$\mathrm{a}=150 \mathrm{m}/{\mathrm{s}}^2$

Positive sign implies direction is towards mean position.

 Case 2 : At 0.05m from mean position

If x=-0.025m,

$\mathrm{a}=100 \mathrm{m}/{\mathrm{s}}^2$

If x=0.075m,

$\mathrm{a}=-100 \mathrm{m}/{\mathrm{s}}^2$