A particle of mass m moves in a one-dimension | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The potential energy is given by $U(x) = -ax^2 + bx^4$.To find the equilibrium positions,we set the first derivative to zero:$\frac{dU}{dx} = -2ax

Vedclass · Accepted Answer

$2 \sqrt{\frac{a}{m}}$

Vedclass · Answer

(B) The potential energy is given by $U(x) = -ax^2 + bx^4$.To find the equilibrium positions,we set the first derivative to zero:$\frac{dU}{dx} = -2ax + 4bx^3 = 0$$2x(2bx^2 - a) = 0$This gives equilibrium points at $x = 0$ and $x = \pm \sqrt{\frac{a}{2b}}$.To find the minima,we check the second derivative: $\frac{d^2U}{dx^2} = -2a + 12bx^2$.At $x = 0$,$\frac{d^2U}{dx^2} = -2a At $x = \pm \sqrt{\frac{a}{2b}}$,$\frac{d^2U}{dx^2} = -2a + 12b(\frac{a}{2b}) = -2a + 6a = 4a > 0$ (stable equilibrium).The effective spring constant $k$ at the minima is $k = \frac{d^2U}{dx^2} = 4a$.The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{k}{m}}$.Substituting $k = 4a$,we get $\omega = \sqrt{\frac{4a}{m}} = 2\sqrt{\frac{a}{m}}$.

$A$ particle of mass $m$ moves in a one-dimensional potential energy $U(x) = -ax^2 + bx^4$,where $a$ and $b$ are positive constants. The angular frequency of small oscillations about the minima of the potential energy is equal to

Similar Questions

$P$ and $Q$ are fixed points in the same plane and a mass $m$ is tied by strings as shown in the figure. If the mass is displaced slightly out of this plane and released,it will oscillate with a time period (given $PQ = 2d$,$PR = QR = L$).

The potential energy of a particle of mass $10 \ g$ as a function of displacement $x$ is $(50 x^2 + 100) \ J$. The frequency of oscillation is

$A$ small spherical ball of radius '$r$' is rolling on a curved surface which is frictionless and has a radius of curvature '$R$'. Its motion is simple harmonic. Then its time period of oscillation is proportional to ($g=$ acceleration due to gravity).

$A$ particle executes linear simple harmonic motion with an amplitude of $2 \ cm$. When the particle is at $1 \ cm$ from the mean position,the magnitude of its velocity is equal to that of its acceleration. Then its time period in seconds is

$A$ particle of mass $m$ is executing oscillations about the origin on the $X-$axis. Its potential energy is $U(x) = k|x|^3$,where $k$ is a positive constant. If the amplitude of oscillation is $a$,then its time period $T$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module