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
The potential energy of the particle is given by
U=−∫Fdx=−∫(−kx+ax3)dx
⇒U=kx22−ax44=x24(2k−ax2).......(1)
From Eq. (1) it follows that U = 0 at two values of x which are x = 0 and x=√2ka. Hence, graphs (b) and (c) are not possible.
Also, U is maximum or minimum at a value of x given by dUdx=0
⇒ddx(kx22−ax44)=0⇒kx−ax3=0
⇒x=√ka
At this value of x, U is maximum if d2Udx2<0.
Now, d2Udx2=ddx(kx−ax3)=k−3ax2
At x=√ka, d2Udx2=k−3aka
=k−3k=−2k, which is negative.
Hence U is maximum at x=√ka.
Hence, graph (a) is also not possible.
Moreover, U is negative for x>√2ka.
Therefore, the correct graph is (d).