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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 of the particle from the origin as F(x)=-kx+ax^(3). Hence k and a are positive constants. For 20. the functional form of the potential energy U(x) of the particle is: |
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Answer»
Also, we know that `F(x)=-(dU)/(dx)` `:.P.E., U=-intF.dx` =`-INT(-kx+ax^3)dx` `U=(kx^2)/(2)-(ax^4)/(4)` This follows that U=0 at two valves of x i.e. `(kx^2)/(2)-(ax^4)/(4)=0` `(x^2)/(2)[k-(ax^2)/(2)]=0` `:.` x=0 and`x=sqrt((2k)/(a))`. This means the graphs (b) and (C) are not possible. Further U is maximum or minimum, when `(dU)/(dx)=0`. i.e. `-(-kx+ax^3)=0` `kx-ax^3=0` `x(k-ax^2)=0` `:.` EITHER x=0 or `x=sqrt(k/a)`. For the value of `sqrt(k/a)=x`, U will be max., Where `(d^2U)/(dx^2)lt0` Now `(d^2U)/(dx^2)=(d)/(dx)(kx-ax^3)=k-3ax^2` =`k-3a(sqrt(k/a))^2=-2k` THUS, U is max. at `x=sqrt(k/a)` and minimum NEGATIVE for `xgtsqrt((2k)/(a))`.Hence graph (a) is also ruled out. |
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