Question

A particle of mass $m$ moves in a one-dimensional potential energy $U\left(x\right)=-a{x}^2+b{x}^4$, where '$a$' and '$b$' are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to

• A. $\pi \sqrt{\frac{a}{2b}}$
• B. $2\sqrt{\frac{a}{m}}$
• C. $\sqrt{\frac{2a}{m}}$
• D. $\sqrt{\frac{a}{2m}}$

Answer 1

The correct option is B $2\sqrt{\frac{a}{m}}$

$\frac{dV\left(x\right)}{dx}=\frac{d}{dx}(-a{x}^2+b{x}^4)$

$=-2ax+4b{x}^3=x(4b{x}^2-2a)=0$

$x_1=0x_2=\sqrt{a/2b}$

$F=-kx\frac{dF}{dx}=k=\frac{d^{2}V\left(x\right)}{d{x}^2}$

$k=\frac{d}{dx}(-2ax+4b{x}^3)=-2a+12b{x}^2$

$k(x=0)=-2a$

$k(x=\sqrt{a/2b})=-2a+6a=4a$

$\omega =\sqrt{\frac{k}{m}}$

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

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