Question

Consider the well-known problem of a cart of mass \(m\) moving along the \(x\) axis attached to a spring (force constant \(k\) ), whose other end is held fixed (Figure 5.2 ). If we ignore the mass of the spring (as we almost always do) then we know that the cart executes simple harmonic motion with angular frequency \(\omega=\sqrt{k / m} .\) Using the Lagrangian approach, you can find the effect of the spring's mass \(M,\) as follows: (a) Assuming that the spring is uniform and stretches uniformly, show that its kinetic energy is \(\frac{1}{6} M \dot{x}^{2} .\) (As usual \(x\) is the extension of the spring from its equilibrium length.) Write down the Lagrangian for the system of cart plus spring. (Note: The potential energy is still \(\frac{1}{2} k x^{2}\).) (b) Write down the Lagrange equation and show that the cart still executes SHM but with angular frequency \(\omega=\sqrt{k /(m+M / 3)} ;\) that is, the effect of the spring's mass \(M\) is just to add \(M / 3\) to the mass of the cart.

Short answer

The spring's mass adds \( \frac{M}{3} \) to the cart's effective mass.

Step-by-step solution

Calculate Kinetic Energy of the Spring

To find the kinetic energy of the spring, assume that it is uniform and stretches uniformly when the cart moves. When the end of the spring connected to the cart moves with velocity \( \dot{x} \), the velocity at a distance \( x' \) from the fixed end of the spring is \( \frac{x'}{x} \dot{x} \). The kinetic energy of an infinitesimal piece of the spring is then \( \frac{1}{2} \left( dx' \right) M \left( \frac{x'}{L} \right) \dot{x}^{2} \), where \( L \) is the equilibrium length of the spring. Integrate this expression from \( 0 \) to \( L \) to obtain the total kinetic energy of the spring: \( \int_{0}^{L} \frac{1}{2} \left( dx' \right) M \left( \frac{x'}{L} \right) \dot{x}^{2} = \frac{1}{6} M \dot{x}^{2} \).

Write the Lagrangian for the System

The Lagrangian \( \mathcal{L} \) for the cart and the spring is the difference between their kinetic and potential energies. The kinetic energy of the cart is \( \frac{1}{2} m \dot{x}^{2} \), and the kinetic energy of the spring, as determined in Step 1, is \( \frac{1}{6} M \dot{x}^{2} \). The potential energy is given as \( \frac{1}{2} k x^{2} \). Thus, the Lagrangian is: \( \mathcal{L} = \frac{1}{2} m \dot{x}^{2} + \frac{1}{6} M \dot{x}^{2} - \frac{1}{2} k x^{2} \).

Derive Equation of Motion Using Lagrange's Equation

Lagrange's equation is \( \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) - \frac{\partial \mathcal{L}}{\partial x} = 0 \). First, compute \( \frac{\partial \mathcal{L}}{\partial \dot{x}} = \left( m + \frac{1}{3} M \right) \dot{x} \). Then, \( \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) = \left( m + \frac{1}{3} M \right) \ddot{x} \). Next, \( \frac{\partial \mathcal{L}}{\partial x} = -kx \). Substitute these into Lagrange's equation: \( \left( m + \frac{1}{3} M \right) \ddot{x} + kx = 0 \).

Simplify the Equation of Motion

The equation from Step 3 represents a simple harmonic oscillator, \( \left( m + \frac{1}{3} M \right) \ddot{x} + kx = 0 \), where \( m + \frac{1}{3} M \) is the effective mass of the system. Comparing with the standard form \( m_{\text{eff}} \ddot{x} + kx = 0 \), we identify the effective mass \( m_{\text{eff}} = m + \frac{1}{3} M \). The angular frequency \( \omega \) is then given by \( \omega = \sqrt{\frac{k}{m_{\text{eff}}}} = \sqrt{\frac{k}{m + \frac{M}{3}}} \).

Key concepts

Simple Harmonic Motion

Simple Harmonic Motion (SHM) is a type of periodic motion where an object moves back and forth through an equilibrium position. In the context of a cart attached to a spring, the motion is controlled by the restoring force of the spring. This force is proportional to the displacement of the cart from its equilibrium position according to Hooke's Law, which is defined as \( F = -kx \), where \( k \) is the spring constant and \( x \) is the displacement.

• The system oscillates because the spring force continuously attempts to return the cart to its original position.
• The motion occurs at an angular frequency \( \omega \), which provides insight into how rapidly the cart oscillates.

Using Lagrangian mechanics, we can determine how the mass of the spring affects the system. Without considering the spring's mass, the angular frequency is \( \omega = \sqrt{\frac{k}{m}} \). However, when the spring's mass \( M \) is considered, the system's behavior changes. It adjusts the effective mass in the system, altering the angular frequency to \( \omega = \sqrt{\frac{k}{m + \frac{M}{3}}} \). This demonstrates that the spring's mass indeed impacts the oscillatory behavior, adding \( \frac{M}{3} \) to the cart's mass.

Kinetic Energy

Kinetic Energy in mechanics refers to the energy that an object possesses due to its motion. For our system of a cart and a spring, both these components have kinetic energy as they move. The kinetic energy \( K \) is represented by the formula \( K = \frac{1}{2} mv^2 \), where \( m \) is mass and \( v \) is velocity.

• For the cart, its kinetic energy is straightforwardly computed as \( \frac{1}{2} m \dot{x}^2 \), where \( \dot{x} \) is its velocity.
• The spring, being uniform, has parts that are moving at different speeds. Its kinetic energy is \( \frac{1}{6} M \dot{x}^{2} \) as derived from integration over its length.

This addition of a spring's kinetic energy shows how energy is distributed in a mechanical system. A weighted portion of the spring’s mass contributes to the system's overall kinetic energy, affecting the system's dynamic behavior.

Analytical Mechanics

Analytical Mechanics, particularly Lagrangian Mechanics, offers powerful tools for solving complex mechanical systems. It focuses on the total energy in systems rather than individual forces and allows for the simplification of the equations that govern motion. The Lagrange approach reformulates Newtonian mechanics, making it easier to handle constraints and conserve energy.
The Lagrangian \( \mathcal{L} \) is defined as the difference between kinetic energy \( T \) and potential energy \( V \):

• \( \mathcal{L} = T - V \)
• For a cart-spring setup, this becomes \( \mathcal{L} = \frac{1}{2} m \dot{x}^{2} + \frac{1}{6} M \dot{x}^{2} - \frac{1}{2} k x^{2} \).

Using the Lagrange equation \( \frac{d}{dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{x}} \right) - \frac{\partial \mathcal{L}}{\partial x} = 0 \), we derive the equations of motion. This results in the familiar form of the simple harmonic oscillator equation with an adjusted effective mass: \( (m + \frac{1}{3} M) \ddot{x} + kx = 0 \). Through the lens of analytical mechanics, we can see the elegance and efficiency in solving for the dynamics of physical systems.