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 ${\nu }_0$. It performs simple harmonic oscillations. The frequency of the simple harmonic motion depends on

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

Answer 1

Answer: (C)

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