WBJEE · Physics · Oscillations
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
- A \(2 \pi \sqrt{\frac{m}{(A p)}}\)
- B \(2 \pi \sqrt{\frac{m}{\left(A p^2\right)}}\)
- C \(2 \pi \sqrt{\frac{\mathrm{m}}{\mathrm{A}}}\)
- D \(\frac{1}{2 \pi} \sqrt{\frac{A p}{m}}\)
Answer & Solution
Correct Answer
(B) \(2 \pi \sqrt{\frac{m}{\left(A p^2\right)}}\)
Step-by-step Solution
Detailed explanation
Hints: \(v_x=A(1-\cos p x)\) \(F=-\frac{d u}{d x}=-A p \sin p x\) For small (x) \(\mathrm{F}=-\mathrm{AP}^2 \mathrm{x}\) \(a=-\frac{A p^2}{m} x \quad a=-\omega^2 x\) \(\omega=\sqrt{\frac{A P^2}{\mathrm{~m}}} \quad \therefore \mathrm{T}=2 \pi \sqrt{\frac{\mathrm{m}}{A p^2}}\)
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