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

Q: 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

a (a) Force on the particle is $F=-\frac{d V}{d x}$ $\Rightarrow \quad F=-\frac{d}{d x}\left(\frac{1}{2} k x^2-V_o \cos \left(\frac{x}{a}\right)\right)$ $=-\left(k x+\frac{V_o}{a} \sin \left(\frac{x}{a}\right)\right)$ As $\quad x << a, \Rightarrow \frac{x}{a} << 1$ So, $\sin \frac{x}{a} \approx \frac{x}{a}$ Hence, $F=-\left(k x+\frac{V_0}{a} \cdot \frac{x}{a}\right)=-\left(k+\frac{V_0}{a^2}\right) \cdot x$ Acceleration $A$ of the particle is $A=\frac{F}{m}=-\frac{1}{m}\left(k+\frac{V_{0}}{a^2}\right) \cdot x$ As, acceleration $A=-\omega^2 x$, we have $c 0=\sqrt{\frac{1}{m}\left(k+\frac{V_0}{a^2}\right)}=\sqrt{\left(\frac{k a^2+V_0}{m a^2}\right)}$ $\therefore$ Time period of oscillation is $T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\left(\frac{m a^2}{k a^2+V_0}\right)}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

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

A$2 \pi \sqrt{\frac{m a^2}{k a^2+V_0}}$
B$2 \pi \sqrt{\frac{m}{k}}$
C$2 \pi \sqrt{\frac{m a^2}{V_0}}$
D$2 \pi \sqrt{\frac{m a^2}{k a^2-V_0}}$

KVPY 2010, Advanced

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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