A particle of mass 5 times 10^{-5} kg is pla | vedclass

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(C) The equation of the parabola is $x^2 = 40y$,which implies $y = \frac{x^2}{40}$.The slope of the tangent at any point $x$ is given by $	an 	heta

Vedclass · Accepted Answer

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

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