Physics Problems & Solutions: Classical Mechanics

A Particle of mass m moves in a one-dimensional potential V(x) = −ax² + bx⁴, where a and b are positive constants. The angular frequency of small oscillations about the minima of the potential is equal to

A. π(a/2b)^1/2
B. π(a/m)^1/2
C. (a/mb)^1/2
D. 2(a/m)^1/2
E. (a/2m)^1/2

(GR9677 #92)

Solution:

The minima of the potential (most probable value x or the equilibrium position of the mass):

$\frac{dV(x)}{dx}=0$

$\begin{aligned} \frac{dV(x)}{dx}&=\frac{d}{dx} (-ax^2+bx^4)\\&=-2ax+4bx^3\\&=x(4bx^2-2a)=0 \\ \\x_1&=0 \\ x_2&=\sqrt{a/2b} \end{aligned}$

Angular Frequency: $\omega=\sqrt{\frac{k}{m}}$

Conservative force:
$\\F=-\nabla{V} =-kx \rightarrow \frac{dF}{dx} = k = \frac{d^2V(x)}{dx^2}$

$\begin{aligned} k=\frac{d}{dx}(-2ax+4bx^3)&=-2a+12bx^2 \\k(x=0)&= -2a \\k(x=\sqrt{a/2b})&=-2a+6a= 4a \end{aligned}$

Thus, the angular frequency about the minima of the potential:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{4a}{m}}=2\sqrt{\frac{a}{m}}$

Answer: D

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