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(C) The equation of the parabola is $x^2 = 40y$,which gives $y = \frac{x^2}{40}$.The slope of the tangent at any point $x$ is $\frac{dy}{dx} = \frac{2

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$\frac{1}{\sqrt{2}}$

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(C) The equation of the parabola is $x^2 = 40y$,which gives $y = \frac{x^2}{40}$.The slope of the tangent at any point $x$ is $\frac{dy}{dx} = \frac{2x}{40} = \frac{x}{20}$.For small displacements,the angle $	heta$ that the tangent makes with the horizontal is small,so $	an 	heta = \frac{dy}{dx} = \frac{x}{20}$.The restoring force acting on the particle is $F = -mg \sin 	heta \approx -mg 	an 	heta$.Substituting $	an 	heta$,we get $F = -mg \left(\frac{x}{20}ight)$.Using Newton's second law,$F = ma$,so $ma = -mg \left(\frac{x}{20}ight)$,which simplifies to $a = -\left(\frac{g}{20}ight)x$.Comparing this with the standard $SHM$ equation $a = -\omega^2 x$,we get $\omega^2 = \frac{g}{20}$.Taking $g = 10 \ m/s^2$,we have $\omega^2 = \frac{10}{20} = \frac{1}{2}$.Therefore,$\omega = \frac{1}{\sqrt{2}} \ rad/s$.

$A$ particle of mass $5 \times 10^{-5} \ kg$ is placed at the lowest point of a smooth parabola $x^2 = 40y$ ($x$ and $y$ in $m$). If it is displaced slightly such that it is constrained to move along the parabola,the angular frequency of oscillation (in $rad/s$) will be approximately:

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