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A particle of mass \mathrm{m} is confined to move in a linear potential given by \mathrm{U}=\mathrm{K} \mathrm{r}, where \mathrm{K} is a constant and r represents the radial distance from the origin. When the particle is in its equilibrium state, it undergoes circular motion with a fixed radius r and angular frequency \omega_0.

If the particle deviates slightly from this circular motion, it will experience small oscillations around its equilibrium position. The angular frequency of these oscillations is denoted as \omega and can be expressed as \omega=\sqrt{n} \omega_0.

Find The value of n, which represents the ratio of the angular frequency of the small oscillations \omega to the angular frequency of the stationary circular motion \omega_0.

Option: 1

1

 


Option: 2

2

 


Option: 3

3

 


Option: 4

4


Answers (1)

best_answer

The total energy is given by E_{\text {total }}=\frac{3 K r}{2}, where \mathrm{K} is the constant associated with the linear potential U=K r.

The angular momentum about the origin is L=\sqrt{m K r^3} \text {, }

and the angular frequency of circular motion is \omega=\sqrt{\frac{K}{m r}}

The effective potential is U_{\text {eff }}=K r+\frac{L^2}{2 m r^2}

Radius r_0 of the stationary circular motion is: 

\left(\frac{\mathrm{dU}_{\mathrm{eff}}}{\mathrm{dr}}\right)_{\mathrm{r}=\mathrm{I}_0}=\mathrm{K}-\frac{\mathrm{L}^2}{\mathrm{mr}_0^3}=0

The radius of the stationary circular motion is r_0=\left(\frac{L^2}{m K}\right)^{\frac{1}{3}}

The second derivative of the effective potential with respect to r is \frac{3 L^2}{m}\left(\frac{m K}{L^2}\right)^{\frac{4}{3}}.

The angular frequency of small radial oscillations about r_0  if it is slightly disturbed from the stationary circular motion is:
\omega_{\mathrm{r}}=\sqrt{\frac{1}{\mathrm{~m}}\left(\frac{\mathrm{d}^2 \mathrm{U}_{\mathrm{eff}}}{\mathrm{dr}^2}\right)_{r=I_{0}}}=\sqrt{\frac{3 \mathrm{~K}}{\mathrm{~m}}\left(\frac{\mathrm{mK}}{\mathrm{L}^2}\right)^{\frac{1}{3}}}=\sqrt{\frac{3 \mathrm{~K}}{\mathrm{mr}_0}}=\sqrt{3} \omega_0,

 , where \omega_o is the angular frequency of the stationary circular motion. 

Therefore, according to the given condition 

\omega=\sqrt{n} \omega_o \text {, we find that } n=3 \text {. }

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