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 $\left(x=0\right)$ 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

Hint: Try to get the equation of acceleration using the second law of Newton mechanics to relate it to the acceleration of particles performing simple harmonic oscillation. For that, first, find the force by differentiating potential energy.

Formula used:
 $F=-{\displaystyle \frac{dU}{dx}}$; Representing force(F) is the (negative)derivative of Potential energy($U$).
$a\left(t\right)=-{\omega }^{2}x\left(t\right)$, where $a(t)$ is acceleration with respect to time, $\omega$ is angular velocity and $x(t)$ is the displacement function.

Complete answer:
Here, a particle is performing a simple harmonic motion in one-dimensional motion which has potential energy at point $x$ from the mean position is $U$.
We have,
Potential energy, $$U=a+b{x}^2$$
Where, $a$ and $b$ are constants
Now Force, $F=-{\displaystyle \frac{dU}{dx}}$
 By putting the value of $U$in the above equation, we get
$$\begin{gathered} F=-{\displaystyle \frac{d\cdot \left(a+b{x}^2\right)}{dx}} \\ F=-\left[{\displaystyle \frac{d}{dx}}\left(a\right)+{\displaystyle \frac{d}{dx}}\left(b{x}^2\right)\right] \end{gathered}$$
Since the derivative of constant is zero, so
${\displaystyle \frac{d}{dx}}\left(a\right)=0$
By substituting this, we have
$\Rightarrow F=0-b\cdot {\displaystyle \frac{d}{dx}}\left(x^2\right)$
$\Rightarrow F=0-b\cdot 2x$ (derivative of $x^n$ is $n{x}^{\left(n-1\right)}$)
$\Rightarrow F=-2bx\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$
We know that, $F=ma$ from second law of newton where,
$m$=mass of particle
$a$=acceleration of particle
So,
$\Rightarrow a={\displaystyle \frac{F}{m}}$
Put the value of $F$that we get in equation $\left(1\right)$
$\Rightarrow a={\displaystyle \frac{F}{m}}=-{\displaystyle \frac{2b}{m}}x\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$
The acceleration of a particle performing simple harmonic motion is given by,
$a\left(t\right)=-{\omega }^{2}x\left(t\right)\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)$
Here, $\omega$ is the angular velocity of a particle.
By comparing $\left(2\right)$and$\left(3\right)$, we get
$$\begin{gathered} \Rightarrow {\omega }^2={\displaystyle \frac{2b}{m}} \\ \Rightarrow \omega =\sqrt{{\displaystyle \frac{2b}{m}}} \end{gathered}$$

$\therefore$ The frequency of simple harmonic motion depends on $b$and $m$.
So the correct option is A.

Note:
One interesting characteristic of the SHM of an object attached to a spring is the angular frequency, and therefore the period and frequency of the motion, depending on only the mass and the force constant, and not on other factors such as the amplitude of the motion.