Question

If the potential energy is minimum then its second derivative is positive .why?​

Answer 1

$\huge\mathmark{hello-dear}$

answer=> Consider the following potentials:

U(x)= x^4
U(x)= x^6-x^4
U(x)= x^4+x^3

All three of these potentials have an equilibrium point at x=0x=0. All three of these potentials are such that the second derivative of U(x)U(x) at this equilibrium point is zero. However, you should convince yourself (perhaps by plotting these potentials) that in the first case the equilibrium is stable, in the second case it is unstable, and in the third case the equilibrium is, as you put it, "stable in one direction but unstable in the other".

The moral is: knowing only that the second derivative is zero tells us nothing about stability. We need to look at higher derivatives if we want to know more.


$<marquee > hope its help you$

Answer 2

if potential energy is min. then K.E. is max.

and 1st derivative of P.E. is -force and 2nd is momentum ,

K.E. is max means speed will decrease so momentum will be negative

and force is negative of dU/de

so 2nd derivative is positive

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