A particle of mass m is located in a one-dimensional potential field- where the potential energy of the particle depends on the coordinate X as U x - U 01-cos- x - Where U 0 and - are constants- Find the time period of small oscillations that the particle performs about the equilibrium position-A- 4 - m - U 0-2B- 2 - m -2 U 0-2C- 2 -m- U 0-2D- - m - U 0-2

The correct option is C 2 π√(m/U_0β^2)U(x) = U_0 (1 cos⁡β x) F=|dU/dx|=β U_0 sin β x f'(0)=|dU/dx|_x 0=U_0 β^2 cos 0^∘=U_0 β^2 T=2π√(m/f'(0))=2π√(m/U_0β^2) ...

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Byju's Answer

Standard XII

Physics

Damped Oscillations

A particle of...

Question

A particle of mass m is located in a one-dimensional potential field, where the potential energy of the particle depends on the coordinate X as $U (x) = U_{0} (1 - c o s β x)$ , Where $U_{0}$ and $β$ are constants. Find the time period of small oscillations that the particle performs about the equilibrium position.

4 π \sqrt{\frac{m}{U_{0} β^{2}}}

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2 π \sqrt{\frac{m}{U_{0} β^{2}}}

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2 π \sqrt{\frac{m}{2 U_{0} β^{2}}}

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π \sqrt{\frac{m}{U_{0} β^{2}}}

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Solution

The correct option is B $2 π \sqrt{\frac{m}{U_{0} β^{2}}}$
$U (x) = U_{0} (1 - c o s β x)$
$F = ∣ ∣ \frac{d U}{d x} ∣ ∣ = β U_{0} s i n β x$
$f^{'} (0) = {∣ ∣ \frac{d U}{d x} ∣ ∣}_{x - 0} = U_{0} β^{2} c o s 0^{\circ} = U_{0} β^{2}$
$T = 2 π \sqrt{\frac{m}{f^{'} (0)}} = 2 π \sqrt{\frac{m}{U_{0} β^{2}}}$

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A particle of mass m is located in a one-dimensional potential field, where the potential energy of the particle depends on the coordinate X as U(x)=U0(1−cosβx), Where U0 and β are constants. Find the time period of small oscillations that the particle performs about the equilibrium position.

The correct option is B 2π√mU0β2U(x)=U0(1−cosβx) F=∣∣dUdx∣∣=βU0 sin βx f′(0)=∣∣dUdx∣∣x−0=U0β2 cos 0∘=U0β2 T=2π√mf′(0)=2π√mU0β2