WBJEE 2024 Physics & Chemistry Paper

To determine the angular frequency of the oscillation of a particle under the given potential energy function, we first need to find the effective force acting on the particle by taking the negative gradient of the potential energy:$F(x)=-\frac{dU(x)}{dx}$Given the potential energy function:$U(x)=-\frac{\alpha x}{x^{2}+\beta^{2}}$ we can compute the force as follows:$F(x)=-\frac{d}{dx}\left(-\frac{\alpha x}{x^{2}+\beta^{2}}\right)$Performing the differentiation using the quotient rule for $\frac{u}{v}$, where $u=\alpha x$ and $v=x^{2}+\beta^{2}$ $\frac{d}{dx}\left(\frac{\alpha x}{x^{2}+\beta^{2}}\right)=\frac{(\alpha)(x^{2}+\beta^{2})-(\alpha x)(2x)}{(x^{2}+\beta^{2})^{2}}$Simplifying the numerator:$\alpha x^{2}+\alpha\beta^{2}-2\alpha x^{2}=\alpha\beta^{2}-\alpha x^{2}$ So the force is:$F(x)=-\frac{\alpha\beta^{2}-\alpha x^{2}}{(x^{2}+\beta^{2})^{2}}$In simple harmonic motion, near the equilibrium position at $x=0$, the potential energy function can be approximated as a quadratic function of $x$. That is:$U(x)\approx\frac{1}{2}kx^{2}$ where $k$ is the effective spring constant. The force near the equilibrium can also be approximated as:$F(x)\approx -kx$ Here, we can determine $k$ by comparing the second derivative of the potential energy function at $x=0$.Taking the second derivative of $U(x)$ with respect to $x$ : $\frac{d^{2}U(x)}{dx^{2}}=\frac{d}{dx}\left(\frac{dU(x)}{dx}\right)$We previously found the first derivative as:$\frac{dU(x)}{dx}=\frac{\alpha\beta^{2}-\alpha x^{2}}{(x^{2}+\beta^{2})^{2}}$Evaluating the second derivative at $x=0$ :$\left.\frac{d^{2}U(x)}{dx^{2}}\right|_{x=0}=\left.\frac{d}{dx}\left(\frac{\alpha\beta^{2}}{(\beta^{2})^{2}}\right)\right|_{x=0}$Since the potential energy is symmetric around $x=0$ and the first derivative at $x=0$ is zero, we get: $\left.\frac{d^{2}U(x)}{dx^{2}}\right|_{x=0}=\left.\frac{d}{dx}\left(\frac{\alpha}{\beta^{2}}\right)\right|_{x=0}$So, the effective spring constant $k$ is:$k=\frac{\alpha}{\beta^{2}}$ The angular frequency $\omega$ of the oscillation for a mass $m$ attached to a spring constant $k$ is given by:$\omega=\sqrt{\frac{k}{m}}$Substituting the value of $k$ : $\omega=\sqrt{\frac{\alpha}{m\beta^{2}}}$This shows that the angular frequency is proportional to:$\sqrt{\frac{\alpha}{m\beta^{3}}}$Thus, the correct option is:Option C: $\sqrt{\frac{\alpha}{m\beta^{3}}}$