1tnks sir.
A particle moves under the influence of the potential V(x) = -Cx^{n}e^{-ax}. Find the frequency of
small oscillations around the equilibrium point.
ans not given please post workin' . NOTHING MORE mentioned in the question.
-
UP 0 DOWN 0 2 2
2 Answers
66The potential energy is U(x)= m V(x).
The general idea is to approximate the potential energy function in the vicinity of an equilibrium point by a parabolic potential energy. To this end, expand the potential energy function about the equilibrium point (say x0) in a Taylor's series:
U(x)=U(x_0)+U'(x_0)(x-x_0)+\dfrac{1}{2!}U''(x_0)(x-x_0)^2+ \ldots
Now since x0 is an equilibrium point, so U'(x0)=0. Further, if the displacement from x0 is small enough, so that (x&ndash x0)3 and higher powers can be neglected, then the potential energy takes the form
U(x)=U(x_0)+\dfrac{1}{2!}U''(x_0)(x-x_0)^2
Shifting the origin to the point (x0, U(x0)), the potential energy takes the form
u(X)=\dfrac{1}{2}U''(x_0)X^2
where the new coordinate X = x-x0 measures the displacements from the equilibrium while u(X) = U(x) &ndash U(x0) is new potential energy form.
Comparing the last equation with the potential energy of a spring mass system, we see that the "spring constant"
k=U''(x_0)
Accordingly, the angular frequency of small oscillations
\omega =\sqrt{\dfrac{k}{m}}=\sqrt{\dfrac{U''(x_0)}{m}}
For the present case,
U(x)=mV(x)=-mCx^ne^{-ax}
Hence, the equilibrium position x0 = na
Hence
U''(x_0)=mCae^{-n}\left(\dfrac{n}{a}\right)^{n-1}
You can now finish things off. [1]