A particle of mass 5 × 10-5 kg is placed at the lowest point of a smooth parabola having the equation 20 x2=y(x, y in m ) . Here y is the vertical height. If it is displaced slightly and it is constrained to move along the parabola, find the angular frequency (in rad / s ) of small oscillations.

Question

A particle of mass 5 × 10-5 kg is placed at the lowest point of a smooth parabola having the equation 20 x2=y(x, y in m ) . Here y is the vertical height. If it is displaced slightly and it is constrained to move along the parabola, find the angular frequency (in rad / s ) of small oscillations. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Potential energy of particle =m g y ∴ Fx=(- partial U/ partial x)=(- partial(20 m g x2)/ partial x)=-40 m g x F propto-x, thus ω2=(40 × 5 × 10-5 × 10/5 × 10-5) ⇒ ω=20

Q. A particle of mass 5×10−5kg is placed at the lowest point of a smooth parabola having the equation 20x2=y(x,y in m). Here y is the vertical height. If it is displaced slightly and it is constrained to move along the parabola, find the angular frequency (in rad/s ) of small oscillations.

Potential energy of particle =mgy ∴Fx​=∂x−∂U​=∂x−∂(20mgx2)​=−40mgx F∝−x, thus ω2=5×10−540×5×10−5×10​ ⇒ω=20

Potential energy of particle $= m g y$
$∴ F_{x} = \frac{- \partial U}{\partial x} = \frac{- \partial ( 20 m g x ^{2} )}{\partial x} = - 40 m gx$
$F \propto - x$ , thus $ω^{2} = \frac{40 \times 5 \times 1 0 ^{- 5} \times 10}{5 \times 1 0 ^{- 5}}$
$\Rightarrow ω = 20$