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 $(x = 0)$ it has initial velocity $ν_0$. 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, $F =-\frac{d U}{d x}$
$= -\frac{d \cdot\left(a+ b x^{2}\right)}{d x}$
$= -\left[\frac{d}{d x}(a)+\frac{d}{d x}\left(b x^{2}\right)\right]$
$\Rightarrow F=0-b \cdot \frac{d}{d x}\left(x^{2}\right)$
$\Rightarrow F=0-b \cdot 2 x=-2 b x$
$\Rightarrow a=\frac{F}{m}=-\frac{2 b}{m} x$ ...(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. $\omega=\sqrt{\frac{2 b}{m}}$
$\omega^{2}=\frac{2 b}{m}$
i.e. the frequency of the simple harmonic motion depends on $b$ and $m$ alone.