A particle of mass $m$ is allowed to oscillate near the point of minima of a vertical parabolic path having the equation $x^2=4ay$, where $x-axis$ is along the horizontal direction and $y-axis$ is along the vertical direction. The angular frequency of small oscillations of the particle is

• A

  $\sqrt{\frac{8g}{a}}$
• B

  $\sqrt{\frac{2g}{a}}$
• C

  $\sqrt{\frac{g}{a}}$
• Correct answer
  D

  $\sqrt{\frac{g}{2a}}$

Solution

$\text{ma}=-\text{mg}sin\theta$

$\text{a}=-g\; sin\theta$  or  $\text{a}=-g\; tan\theta$           ...(i)

(as $\theta$ is small)

Now, ${\text{x}}^2=4\text{ay}$

$\therefore \frac{\text{dy}}{\text{dx}}=\frac{\text{x}}{2\text{a}}$

$\therefore \text{a}=-\text{g}\frac{\text{x}}{2\text{a}}$

$-{\omega }^2\text{x}=-\frac{\text{gx}}{2\text{a}}$

$\omega =\sqrt{\frac{\text{g}}{2\text{a}}}$

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