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 $V_0$ . It 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

Answer: (A)

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