A particle of mass m is located in a one-dime | vedclass

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Question

(A) The potential energy is given by $V(x) = A(1 - \cos px)$.The force acting on the particle is $F = -\frac{dV}{dx} = -\frac{d}{dx} [A(1 - \cos px)]

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$2\pi \sqrt{\frac{m}{Ap^2}}$

Vedclass · Answer

(A) The potential energy is given by $V(x) = A(1 - \cos px)$.The force acting on the particle is $F = -\frac{dV}{dx} = -\frac{d}{dx} [A(1 - \cos px)] = -Ap \sin px$.For small oscillations,the displacement $x$ is very small,so $\sin px \approx px$.Substituting this into the force equation: $F = -Ap(px) = -(Ap^2)x$.This is the equation for simple harmonic motion $F = -kx$,where the effective spring constant $k = Ap^2$.The time period $T$ of small oscillations is given by $T = 2\pi \sqrt{\frac{m}{k}}$.Substituting $k = Ap^2$,we get $T = 2\pi \sqrt{\frac{m}{Ap^2}}$.

$A$ particle of mass $m$ is located in a one-dimensional potential field where the potential energy is given by: $V(x) = A(1 - \cos px)$,where $A$ and $p$ are constants. The period of small oscillations of the particle is:

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