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(A) The force on the particle is given by $F = -\frac{dV}{dx}$.$F = -\frac{d}{dx} \left( \frac{1}{2} k x^2 - V_0 \cos \left( \frac{x}{a} \right) \righ

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$2 \pi \sqrt{\frac{m a^2}{k a^2+V_0}}$

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(A) The force on the particle is given by $F = -\frac{dV}{dx}$.$F = -\frac{d}{dx} \left( \frac{1}{2} k x^2 - V_0 \cos \left( \frac{x}{a} ight) ight) = -\left( kx + \frac{V_0}{a} \sin \left( \frac{x}{a} ight) ight)$.Since the amplitude of oscillation is much smaller than $a$,we have $x \ll a$,which implies $\frac{x}{a} \ll 1$. Using the approximation $\sin 	heta \approx 	heta$ for small $	heta$,we get $\sin \left( \frac{x}{a} ight) \approx \frac{x}{a}$.Substituting this into the force equation: $F \approx -\left( kx + \frac{V_0}{a} \cdot \frac{x}{a} ight) = -\left( k + \frac{V_0}{a^2} ight) x$.This is in the form of $F = -K_{eff} x$,where $K_{eff} = k + \frac{V_0}{a^2}$.The angular frequency $\omega$ is given by $\omega = \sqrt{\frac{K_{eff}}{m}} = \sqrt{\frac{k + \frac{V_0}{a^2}}{m}} = \sqrt{\frac{k a^2 + V_0}{m a^2}}$.The time period $T$ is $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{m a^2}{k a^2 + V_0}}$.

$A$ particle of mass $m$ undergoes oscillations about $x=0$ in a potential given by $V(x) = \frac{1}{2} k x^2 - V_0 \cos \left(\frac{x}{a}\right)$,where $V_0, k, a$ are constants. If the amplitude of oscillation is much smaller than $a$,the time period is given by

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