Q122 / 155 · WBJEE 2024 · Physics

A particle of mass ' $m$ ' moves in one dimension under the action of a conservative force whose potential energy has the form of $U(x)=-\frac{\alpha x}{x^2+\beta^2}$ where $\alpha$ and $\beta$ are dimensional parameters. The angular frequency of the oscillation is proportional to

  • A
    $\sqrt{\frac{\alpha^3}{m \beta^4}}$
  • B
    $\sqrt{\frac{\alpha}{m \beta^4}}$
  • Correct answer
    C
    $\sqrt{\frac{\alpha}{m \beta^3}}$
  • D
    $\sqrt{\frac{\alpha}{m \beta^6}}$

Solution

$$ \begin{aligned} \text { Hint: } U & =-\frac{\alpha x}{x^2+\beta^2} \\ \text { here, }[\beta] & =[L] \\ {[\alpha] } & =\left[M^3 T^{-2}\right] \end{aligned} $$ only option (C) has dimension of $\frac{1}{\mathrm{~T}}$

Hence, the correct answer is option: 'C'.

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