A particle of mass m is executing oscillation | vedclass

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

Question

(A) The potential energy is given by $U(x) = k|x|^3$.The force acting on the particle is $F = -\frac{dU}{dx} = -3k|x|^2 	ext{sgn}(x)$.For a particle

Vedclass · Accepted Answer

Proportional to $\frac{1}{\sqrt{a}}$

Vedclass · Answer

(A) The potential energy is given by $U(x) = k|x|^3$.The force acting on the particle is $F = -\frac{dU}{dx} = -3k|x|^2 	ext{sgn}(x)$.For a particle of mass $m$ oscillating with amplitude $a$,the total energy $E$ is conserved and is equal to the potential energy at the extreme position $(x = a)$:$E = U(a) = ka^3$.At any position $x$,the energy is $E = \frac{1}{2}mv^2 + k|x|^3 = ka^3$.Thus,$v = \frac{dx}{dt} = \sqrt{\frac{2k}{m}(a^3 - |x|^3)}$.The time period $T$ is given by $T = 4 \int_{0}^{a} \frac{dx}{v} = 4 \int_{0}^{a} \frac{dx}{\sqrt{\frac{2k}{m}(a^3 - x^3)}}$.Let $x = ay$,then $dx = a dy$. When $x=0, y=0$ and when $x=a, y=1$.$T = 4 \sqrt{\frac{m}{2k}} \int_{0}^{1} \frac{a dy}{\sqrt{a^3(1 - y^3)}} = 4 \sqrt{\frac{m}{2ka}} \int_{0}^{1} \frac{dy}{\sqrt{1 - y^3}}$.Since the integral is a constant,$T \propto \frac{1}{\sqrt{a}}$.

$A$ 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:

Similar Questions

$A$ bead of mass $m$ is attached to the mid-point of a taut,weightless string of length $l$ and placed on a frictionless horizontal table. Under a small transverse displacement $x$,as shown in the figure,if the tension in the string is $T$,then the frequency of oscillation is:

Under the influence of force $F_1$,a body oscillates with a period $T_1$,and due to another force $F_2$,the body oscillates with a period $T_2$. If both forces act simultaneously,what is the resultant period? (Consider the displacement is the same in all three cases.)

$A$ uniform stick of mass $M$ and length $L$ is pivoted at its centre. Its ends are tied to two springs each of force constant $K$. In the position shown in the figure,the springs are in their natural length. When the stick is displaced through a small angle $\theta$ and released,the stick:

The potential energy of a particle of mass $1\,kg$ in motion along the $x-$axis is given by $U = 4(1 - \cos 2x)\,J,$ where $x$ is in $meters$. The period of small oscillations (in $sec$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module