Question

A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction, which varies with the distance x of the particle from the origin as f(x) = kx + ax3. Here, k and a are positive constants. For x 0, the functional form of the potential energy u(x) of the particle is

Answer 1

Answer:

we are given:-

$f = kx + a {x}^{3}$

and f=-du/dx

so,

$- du = (kx + a {x}^{3} ).dx$

and after indefinite integration:-

$- u = k \frac{ {x}^{2} }{2} + a \frac{ {x}^{4} }{4}$

so finally we have:-

$u = - k \frac{ {x}^{2} }{2} - a \frac{ {x}^{4} }{4}$

so from the above equations it can be concluded that the particle will be at equilibrium at X=0 as the force will be zero(it can known from the equation of force given in the question).

but the particle will be at Unstable equilibrium at X=0 because it is known that the particles have a tendency of having least energy.

(if the particle is deflected to its right or left the particle's potential energy will decrease as -du/dx (which is force ) is zero at X=0 and it shows that the particle have the maximum energy at X=0).

dy/dx=0 means that the function have it's Maxima or minima value at that point.

follow if it helped.

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