The potential energy of a particle of mass m | vedclass

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

Question

(A) Given the potential energy $U(x) = U_0(1 - \cos ax)$.For small oscillations,we can approximate $\cos ax \approx 1 - \frac{(ax)^2}{2}$.Thus,$U(x) \

Vedclass · Accepted Answer

$2\pi \sqrt{\frac{m}{a^2 U_0}}$

Vedclass · Answer

(A) Given the potential energy $U(x) = U_0(1 - \cos ax)$.For small oscillations,we can approximate $\cos ax \approx 1 - \frac{(ax)^2}{2}$.Thus,$U(x) \approx U_0(1 - (1 - \frac{a^2 x^2}{2})) = \frac{1}{2} U_0 a^2 x^2$.Comparing this with the potential energy of a simple harmonic oscillator $U = \frac{1}{2} k x^2$,we identify the effective spring constant $k = U_0 a^2$.The angular frequency of oscillation is $\omega_0 = \sqrt{\frac{k}{m}} = \sqrt{\frac{U_0 a^2}{m}} = a \sqrt{\frac{U_0}{m}}$.The time period $T$ is given by $T = \frac{2\pi}{\omega_0} = \frac{2\pi}{a} \sqrt{\frac{m}{U_0}} = 2\pi \sqrt{\frac{m}{a^2 U_0}}$.

The potential energy of a particle of mass $m$ situated in a unidimensional potential field varies as $U(x) = U_0(1 - \cos ax)$,where $U_0$ and $a$ are constants. The time period of small oscillations of the particle about the mean position is

Similar Questions

$A$ block of mass $M$ rests on a piston executing $S.H.M.$ with a period of $1 \,s$. The amplitude of oscillations, such that the mass is separated from the piston, is (acceleration due to gravity, $g=10 \,m/s^2$, $\pi^2=10$)

$A$ mass on a spring oscillates up and down. As the mass moves downward from its highest point to its equilibrium point:

$A$ spring is stretched by $5 \,\,cm$ by a force of $10 \,\,N$. The time period of the oscillations when a mass of $2 \,\,kg$ is suspended by it is: (in $s$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module