Question

Potential energy of a particle executing SHM along $X-$ axis is given by $U=10+2x+(x-3{)}^2\text{J}$. Find the angular frequency of SHM, if mass of the particle is $3\text{kg.}$

• A. $\sqrt{\frac{2}{3}}\text{rad/sec}$
• B. $\sqrt{\frac{3}{2}}\text{rad/sec}$
• C. $\frac{2}{3}\text{rad/sec}$
• D. $\frac{3}{2}\text{rad/sec}$

Answer 1

The correct option is A $\sqrt{\frac{2}{3}}\text{rad/sec}$

Given that
$U=10+2x+(x-3{)}^2$

Since, potential energy is only function of $x$ we can write that,
$F=-\frac{dU}{dx}$
$\Rightarrow F=-\frac{d}{dx}(10+2x+(x-3{)}^2)=-[0+2+2(x-3)]=-2x+4$
$\Rightarrow F=-2(x-2)$

Comparing the above equation with $F=-k(x-x_0),$ we get,
$k=2\text{N/m}$

Given,
Mass of the particle $\left(m\right)=3\text{kg}$
So, angular frequency of the particle executing SHM is given by $\omega =\sqrt{\frac{k}{m}}$
From the given data, we get
$\omega =\sqrt{\frac{2}{3}}\text{rad/sec}$
Thus, option (a) is the correct answer.