Question

Consider one dimensional motion of a particle of mass $m$. It has potential energy $U=a+b{x}^2$, where $a$ and $b$ are positive constants. At origin $(x=0)$ it has initial velocity $v_0$. If performs simple harmonic oscillations. The frequency of the simple harmonic motion depends on :

• A. $b$ and $m$ alone
• B. $b,a$ and $m$ alone
• C. $b$ alone
• D. $b$ and $a$ alone

Answer 1

The correct option is A $b$ and $m$ alone

Given,
Potential energy, $U=a+b{x}^2$
Force, $F=-\frac{dU}{dx}$
$=-\frac{d\cdot (a+b{x}^2)}{dx}$
$=-\left[\frac{d}{dx}\right(a)+\frac{d}{dx}(b{x}^2\left)\right]$
$\Rightarrow F=0-b\cdot \frac{d}{dx}\left(x^2\right)$
$\Rightarrow F=0-b\cdot 2x=-2bx$
$\Rightarrow a=\frac{F}{m}=-\frac{2b}{m}x.....\left(i\right)$
Comparing Eq. (i) with the standard relation of displacement in case of simple harmonic motion.
i.e. $a=\frac{F}{m}=-{\omega }^{2}x$
i.e. ${\omega }^2=\frac{2b}{m}$
$\omega =\sqrt{\frac{2b}{m}}$
i.e. the frequency of the simple harmonic motion depends on $b$ and $m$ alone.