Question

The potential energy of a particle oscillating along $X-$ axis is given as $U=15+4x+(2x-3{)}^2$, where $U$ is in Joules and $x$ in metres. Total mechanical energy of the particle is $36\text{J}$. Mass of particle is $2\text{kg}$. Then,

• A. Angular frequency of SHM will be $2\text{rad/sec}$
• B. Kinetic Energy at mean position $=16\text{J}$
• C. Amplitude of SHM is $2\text{m}$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A Angular frequency of SHM will be $2\text{rad/sec}$
B Kinetic Energy at mean position $=16\text{J}$
C Amplitude of SHM is $2\text{m}$

Given that
$U=15+4x+(2x-3{)}^2$
Since, the potential energy is only a function of $x,$ for conservative forces,

$F=\frac{-dU}{dx}$
$F=-\frac{d}{dx}(15+4x+(2x-3{)}^2)=-(0+4+2(2x-3)\times 2)=-8x+8$
$\Rightarrow F=-8(x-1)$

Comparing this with $F=-k(x-x_0)$ we get, $k=8\text{N/m}$

Given,
Mass of the particle $\left(m\right)=2\text{kg}$
So, the Angular frequency of the particle $\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{8}{2}}=2\text{rad/s}$

At mean position , $F=0$
$\Rightarrow -8(x-1)=0\Rightarrow x=1\text{m}$


Potential energy at mean position (i.e) $x=1\text{m}$
$U_{mean}=15+4x+(2x-3{)}^2=15+(4)+(2\times 1-3{)}^2=20\text{J}$
Kinetic energy at mean position is given by
$K_{mean}=E-U_{mean}=36-20=16\text{J}$
We know that, kinetic energy at mean position is given by
$K_{mean}=\frac{1}{2}m{\omega }^2{A}^2$
$\Rightarrow 16=\frac{1}{2}m{\omega }^2{A}^2$

From the given data,
we can write that,
$16=\frac{1}{2}\times 2\times (2{)}^2\times A^2$
$\Rightarrow A^2=4\Rightarrow A=2\text{m}$
Thus, options (a) , (b) and (c) are the correct answers.