Question

A particle is placed at the lowest point of a smooth wire frame in the shape of a parabola, lying in the vertical $xy-$ plane having equation $x^2=5y$, where $(x,y)$ are in $\text{metres}$. After slight displacement, the particle is set free to move. Find the angular frequency of oscillation $\text{in rad/s}$.
[Take $g=10{\text{m/s}}^2)$]

• A. $2\text{rad/s}$
• B. $4\text{rad/s}$
• C. $6\text{rad/s}$
• D. $8\text{rad/s}$

Answer 1

The correct option is A $2\text{rad/s}$



Given, $y=\frac{x^2}{5}$
Differentiating with respect to ${}^{\prime }{x}^{\prime }$ on both sides we get,

$\frac{dy}{dx}=0.4x$

On drawing the tangent at the position of particle, we get $\tan \theta$ as the slope.
$\because \text{slope}=\tan \theta =\frac{dy}{dx}$ , at the given position on the wire, we get $\tan \theta =0.4x$

The value of $\theta$ will be very small, as the particle is given a slight displacement from the lowest position.
$\therefore \tan \theta \simeq \sin \theta \simeq \theta$
Or, we can say that, $\sin \theta \simeq 0.4x$

From figure we can deduce that, direction of net restoring force $\left(mg\sin \theta \right)$ is opposite to the direction of displacement of particle.
Hence,
$F_{\text{restoring}}=-mg\sin \theta$
$\Rightarrow ma=-mg\sin \theta$
$\Rightarrow a=-10\left(0.4x\right)=-4x$
Comparing this with $a=-{\omega }^{2}x$, we have
${\omega }^2=4\Rightarrow \omega =2\text{rad/s}$

Thus, option (a) is the correct answer.