Question

Consider one dimensional motion of a particle of mass $m$ . It has potential energy $U=a+bx^{2}$ , where $a$ and $b$ are positive constants. At origin $\left(\right.x=0\left.\right)$ 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, $U=a+bx^{2}$
Force, $F=-\frac{d U}{d x}$
$=-\frac{d . \left(\right. a + b x^{2} \left.\right)}{d x}$
$=-\left[\frac{d}{d x} \left(a\right) + \frac{d}{d x} \left(\right. b x^{2} \left.\right)\right]$
$\Rightarrow \, F=0-b.\frac{d}{d x}\left(\right.x^{2}\left.\right)$
$\Rightarrow \, F=0-b.2x=-2bx$
$\Rightarrow \, a=\frac{F}{m}=-\frac{2 b}{m}x$ .......(i)
Compare equation. (i) with the standard relation of displacement in case of simple harmonic motion.
i.e., $a=\frac{F}{2 m}=-\omega ^{2}x$
i.e., $\omega ^{2}=\frac{2 b}{m}$
$\omega =\sqrt{\frac{2 b}{m}}$
i.e., the frequency of the simple harmonic motion depends on $b$ and $m$ alone.