Question

The potential energy of a particle executing SHM along the x-axis is given by $U=U_0-U_0\cos ax$. What is the period of oscillation?

• A. $2\pi \sqrt{\frac{ma}{U_o}}$
• B. $2\pi \sqrt{\frac{U_o}{ma}}$
• C. $\frac{2\pi }{a}\sqrt{\frac{m}{U_o}}$
• D. $2\pi \sqrt{\frac{m}{a{U}_o}}$

Answer 1

Answer: (C)

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

The potential energy of a particle executing along the x-axis is given by,

$U=U_0-U_{0}cosax$

$U=U_0(1-cosax)$

Force, $F=-\frac{dU}{dx}=-a{U}_{0}sinax$. . . . . .(1)

for small $x$, $sinax=ax$

$m{a}^{\prime }=-a^2{U}_{0}x$

acceleration, $a^{\prime }=\frac{a^2{U}_0}{m}x$

From the above equation,

${\omega }^2=\frac{a^2{U}_0}{m}$ ($a^{\prime }=-{\omega }^{2}x$)

$T=\frac{2\pi }{a}\sqrt{\frac{m}{U_0}}$

The correct option is C.