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A particle of mass \(m\) is located in a one dimensional potential field where potential energy is given by : \(\mathrm{V}(\mathrm{x})=\mathrm{A}(1-\cos \mathrm{px})\), where \(\mathrm{A}\) and \(\mathrm{p}\) are constants. The period of small oscillations of the particle is
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The correct answer is:
\(2 \pi \sqrt{\frac{m}{\left(A p^2\right)}}\)
Hints: \(v_x=A(1-\cos p x)\)
\(F=-\frac{d u}{d x}=-A p \sin p x\)
For small (x)
\(\mathrm{F}=-\mathrm{AP}^2 \mathrm{x}\)
\(a=-\frac{A p^2}{m} x \quad a=-\omega^2 x\)
\(\omega=\sqrt{\frac{A P^2}{\mathrm{~m}}} \quad \therefore \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{A p^2}}\)
\(F=-\frac{d u}{d x}=-A p \sin p x\)
For small (x)
\(\mathrm{F}=-\mathrm{AP}^2 \mathrm{x}\)
\(a=-\frac{A p^2}{m} x \quad a=-\omega^2 x\)
\(\omega=\sqrt{\frac{A P^2}{\mathrm{~m}}} \quad \therefore \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{A p^2}}\)
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