Question

The potential energy of a one-dimensional mass \(m\) at a distance \(r\) from the origin is $$U(r)=U_{0}\left(\frac{r}{R}+\lambda^{2} \frac{R}{r}\right)$$ for \(0 < r < \infty,\) with \(U_{\mathrm{o}}, R,\) and \(\lambda\) all positive constants. Find the equilibrium position \(r_{\mathrm{o}} .\) Let \(x\) be the distance from equilibrium and show that, for small \(x\), the PE has the form \(U=\) const \(+\frac{1}{2} k x^{2}\). What is the angular frequency of small oscillations?

Short answer

Equilibrium position: \( r_0 = \lambda R \); Angular frequency: \( \omega = \sqrt{\frac{2U_0}{\lambda R^2 m}} \).

Step-by-step solution

Find the Equilibrium Position

To find the equilibrium position \( r_0 \), we need to find the point where the potential energy \( U(r) \) is at a minimum. To do this, differentiate \( U(r) \) with respect to \( r \) and set the derivative equal to zero: \[\frac{dU}{dr} = U_0\left(\frac{1}{R} - \lambda^2 \frac{R}{r^2}\right) = 0.\]Solving the equation, we find that\[r_0 = \lambda R.\]

Expand U(r) Around the Equilibrium Point

Next, we perform a Taylor expansion of \( U(r) \) around the equilibrium point \( r_0 \) to find the form \( U = \text{const} + \frac{1}{2} k x^2 \). Let \( x = r - r_0 \), then \( r = r_0 + x \). The linear term in the expansion is zero because \( r_0 \) is an equilibrium point, thus,\[U(r) \approx U(r_0) + \frac{1}{2}U''(r_0)x^2.\]Calculate the second derivative from\[\frac{d^2U}{dr^2} = U_0\left(0 + 2\lambda^2 \frac{R}{r^3}\right).\] Evaluate at \( r = r_0 = \lambda R \):\[\frac{d^2U}{dr^2}\bigg|_{r=r_0} = \frac{2U_0}{\lambda R^2}.\]

Identify k and Write the Potential Energy

Compare the second order term to the form \( \frac{1}{2}kx^2 \). Thus, \( k = \frac{2U_0}{\lambda R^2} \),so the potential energy can be written as\[U = \text{const} + \frac{1}{2} \cdot \frac{2U_0}{\lambda R^2} x^2.\]

Determine the Angular Frequency

The formula for the angular frequency \( \omega \) of small oscillations in a potential energy function of form \( \frac{1}{2}kx^2 \) is given by\[\omega = \sqrt{\frac{k}{m}}.\]Substitute the value of \( k \):\[\omega = \sqrt{\frac{2U_0}{\lambda R^2 m}}.\]

Key concepts

Potential Energy

In classical mechanics, potential energy (U(r) ) represents the stored energy of a system due to its position or configuration. For this exercise, the potential energy of a mass at a distance (r) from the origin is given by:
\[U(r)=U_{0}\left(\frac{r}{R}+\lambda^{2} \frac{R}{r}\right)\]
Here, U_{0}, R,and \lambda are positive constants, each contributing to how the potential energy changes with distance. The potential energy function combines two terms:

• \(\frac{r}{R}\) - which increases linearly with distance.
• \(\lambda^{2} \frac{R}{r}\) - which decreases with distance.

The equilibrium position, where the net force on the system is zero, can be found by setting the derivative of U(r) with respect to (r) to zero. This helps to identify where the potential energy is at a minimum, indicating a stable position for the mass.

Angular Frequency

Angular frequency (\omega) characterizes how quickly a system oscillates around an equilibrium position. In this context, it's associated with the potential energy expression \(\frac{1}{2} k x^{2}\), where (k) is the stiffness or spring constant of the system. The angular frequency is given by:
\[\omega = \sqrt{\frac{k}{m}}\]
This formula shows that angular frequency depends on both the stiffness of the potential energy and the mass:

• A higher stiffness (k) results in faster oscillations.
• A larger mass (m) slows down the oscillations.

By determining (k) from the potential energy function, you can find how the system oscillates. This characteristic is particularly useful in understanding small oscillations around equilibrium.

Taylor Expansion

The Taylor expansion is a mathematical technique used to approximate a function near a specific point. In this exercise, we apply it to the potential energy function around the equilibrium point (r_{0}). The potential energy for small displacements (x) from equilibrium can be written as:
\[U(r) \approx U(r_{0}) + \frac{1}{2}U''(r_{0})x^{2}\]
Here, the first derivative is zero at equilibrium, so only the second-order term matters for small oscillations:

• (x) represents the displacement from equilibrium (r - r_{0}).
• U''(r_{0}) is the second derivative evaluated at (r_{0}), which relates to the curvature of the potential energy function.

This expansion simplifies the potential energy to a form resembling a simple harmonic oscillator, making it easier to analyze small oscillations.

Small Oscillations

Small oscillations refer to the back-and-forth movements of a system around its equilibrium position. For such oscillations, the potential energy simplifies to:
\[U = \text{constant} + \frac{1}{2} k x^{2}\]
This form is similar to the potential energy of a simple harmonic oscillator. The key characteristics of small oscillations include:

• They occur near the equilibrium position (r_{0}).
• The linear restoring force leads to oscillations described by Hooke's law.
• Oscillations are simple and predictable, making calculations straightforward for small displacements.

Understanding small oscillations helps in analyzing many physical systems, as it allows you to model and predict the behavior of the system in a quiet range around its equilibrium.