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A small cart (mass \(m\) ) is mounted on rails inside a large cart. The two are attached by a spring (force constant \(k\) ) in such a way that the small cart is in equilibrium at the midpoint of the large. The distance of the small cart from its equilibrium is denoted \(x\) and that of the large one from a fixed point on the ground is \(X,\) as shown in Figure \(7.13 .\) The large cart is now forced to oscillate such that \(X=A \cos \omega t,\) with both \(A\) and \(\omega\) fixed. Set up the Lagrangian for the motion of the small cart and show that the Lagrange equation has the form $$\ddot{x}+\omega_{0}^{2} x=B \cos \omega t$$ where \(\omega_{\mathrm{o}}\) is the natural frequency \(\omega_{\mathrm{o}}=\sqrt{k / m}\) and \(B\) is a constant. This is the form assumed in Section 5.5, Equation (5.57), for driven oscillations (except that we are here ignoring damping). Thus the system described here would be one way to realize the motion discussed there. (We could fill the large cart with molasses to provide some damping.)

Short Answer

Expert verified
The Lagrange equation is \( \ddot{x} + \omega_0^2 x = B \cos \omega t \), with \( \omega_0 = \sqrt{\frac{k}{m}} \) and \( B = A\omega^2 \).

Step by step solution

01

Identify the System and Coordinates

We must start by identifying the coordinates that describe the motion of the system. The coordinates given are the displacement of the small cart from its equilibrium position, denoted as \( x \), and the displacement of the large cart from a fixed point, \( X = A \cos \omega t \). We assume that the large cart moves according to this prescribed motion while the small cart moves due to the spring force.
02

Write the Lagrangian Function

The Lagrangian \( L \) is defined as the difference between the kinetic energy \( T \) and the potential energy \( U \) of the system. The kinetic energy of the small cart is \( T = \frac{1}{2}m(v_x)^2 \), where \( v_x = \frac{d}{dt}(x+X) = \dot{x} - A\omega \sin \omega t \). The potential energy stored in the spring is \( U = \frac{1}{2}kx^2 \). Thus, the Lagrangian is: \[ L = \frac{1}{2}m(\dot{x} - A\omega \sin \omega t)^2 - \frac{1}{2}kx^2. \]
03

Expand and Simplify the Lagrangian

Expanding the kinetic energy term in the Lagrangian, we get: \( T = \frac{1}{2}m(\dot{x}^2 - 2\dot{x}A\omega \sin \omega t + A^2\omega^2 \sin^2 \omega t) \). Substituting this into the Lagrangian and simplifying, we obtain: \[ L = \frac{1}{2}m\dot{x}^2 - mA\omega \dot{x} \sin \omega t + \frac{1}{2}mA^2\omega^2 \sin^2 \omega t - \frac{1}{2}kx^2. \]
04

Write the Euler-Lagrange Equation

The Euler-Lagrange equation is \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) - \frac{\partial L}{\partial x} = 0 \). We need to evaluate these partial derivatives. Calculating \( \frac{\partial L}{\partial \dot{x}} = m\dot{x} - mA\omega \sin \omega t \) and \( \frac{\partial L}{\partial x} = -kx \), and substituting them into the equation gives us: \[ \frac{d}{dt}(m\dot{x} - mA\omega \sin \omega t) + kx = 0. \]
05

Simplify the Euler-Lagrange Equation

Take the time derivative of \( \frac{\partial L}{\partial \dot{x}} \): \( \frac{d}{dt}(m\dot{x} - mA\omega \sin \omega t) = m\ddot{x} - mA\omega^2 \cos \omega t \). Substituting back into the Euler-Lagrange equation, we have: \[ m\ddot{x} - mA\omega^2 \cos \omega t + kx = 0. \] Rearranging terms, we get: \[ \ddot{x} + \frac{k}{m}x = A\omega^2 \cos \omega t. \]
06

Identify Constants \(\omega_0\) and \(B\)

Recognize that the constant term \( \frac{k}{m} \) represents the natural frequency squared, \( \omega_0^2 = \frac{k}{m} \). The forcing term \( A\omega^2 \cos \omega t \) is equivalent to \( B \cos \omega t \), where \( B = A\omega^2 \). The equation thus takes the form: \[ \ddot{x} + \omega_0^2 x = B \cos \omega t, \] where \( \omega_0 = \sqrt{\frac{k}{m}} \) and \( B = A\omega^2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Driven Oscillations
Driven oscillations describe a system where an external force drives the movement of an object. In the context of the small cart inside the large cart, the system displays driven oscillations because the large cart moves owing to an external periodic force. This motion is given by the equation, \( X = A \cos \omega t \), where \( A \) is the amplitude of oscillation and \( \omega \) is the angular frequency.
The external force causes the small cart, attached to the large cart by a spring, to oscillate driven by both this external motion and the spring's restoring force. Since the motion is influenced by this external periodic drive, it's a typical example of what we call a driven oscillation. This can lead to scenarios where the driven frequency aligns with the natural frequency of the system, resulting in a phenomenon known as resonance, though damping is not considered here. Understanding driven oscillations helps us in designing mechanical systems and predicting their behavior under different driving forces.
  • Driven oscillations can cause resonance, increasing amplitude significantly.
  • The driving force has the form of \( B \cos \omega t \), making the oscillation predictable and periodic.
  • Applications are vast, from electronic circuits to swing sets, reflecting the importance of this concept in engineering and physics.
Euler-Lagrange Equation
To understand and predict the motion of the small cart, we utilize the Euler-Lagrange equation, a fundamental principle in Lagrangian mechanics. The equation provides a way to derive the equations of motion for a system. It's defined generally as:\[\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{x}} \right) - \frac{\partial L}{\partial x} = 0\]where \( L \) is the Lagrangian, a function that summarizes the dynamics of the system. To apply this equation, we first calculate the partial derivatives of the Lagrangian, then substitute them into the Euler-Lagrange equation.
Subsequently, due to the external driving force, the small cart's motion equation is transformed into:\[\ddot{x} + \omega_0^2 x = B \cos \omega t\]This equation reflects the effects of both the spring force (represented by the term \( \omega_0^2 x \)) and the external driving force \( B \cos \omega t \). It's crucial to remember that the Euler-Lagrange equation offers a robust framework, not just for this mechanical system, but for any system where the potential energy can be expressed in terms of generalized coordinates.
  • The Euler-Lagrange equation is essential to finding equations of motion efficiently.
  • It incorporates both kinetic and potential energies via the Lagrangian.
  • Directly ties to how systems evolve over time in classical mechanics.
Natural Frequency
The natural frequency of an oscillating system is a critical concept, often denoted as \( \omega_0 \), representing the rate at which the system would oscillate if it were not damped or driven by an external force. In our exercise, the small cart's natural frequency is defined by \( \omega_0 = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass of the small cart.
Understanding the natural frequency is vital for predicting how the system responds to various forces and determining the resonance conditions. If a system is driven at a frequency that matches its natural frequency, the amplitude of the oscillations may increase significantly, causing potentially destructive resonance.
Natural frequencies are incredibly important across various fields, from designing buildings that can withstand earthquakes to ensuring electronic circuits function effectively without unintended interference.
  • Natural frequency is a defining characteristic of oscillations.
  • Critical for assessing stability and resonance in systems.
  • Impacts design and analysis in engineering and technology fields.

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Most popular questions from this chapter

A mass \(m_{1}\) rests on a frictionless horizontal table and is attached to a massless string. The string runs horizontally to the edge of the table, where it passes over a massless, frictionless pulley and then hangs vertically down. A second mass \(m_{2}\) is now attached to the bottom end of the string. Write down the Lagrangian for the system. Find the Lagrange equation of motion, and solve it for the acceleration of the blocks. For your generalized coordinate, use the distance \(x\) of the second mass below the tabletop.

Write down the Lagrangian for a one-dimensional particle moving along the \(x\) axis and subject to a force \(F=-k x\) (with \(k\) positive). Find the Lagrange equation of motion and solve it.

Two equal masses, \(m_{1}=m_{2}=m,\) are joined by a massless string of length \(L\) that passes through a hole in a frictionless horizontal table. The first mass slides on the table while the second hangs below the table and moves up and down in a vertical line. (a) Assuming the string remains taut, write down the Lagrangian for the system in terms of the polar coordinates \((r, \phi)\) of the mass on the table. (b) Find the two Lagrange equations and interpret the \(\phi\) equation in terms of the angular momentum \(\ell\) of the first mass. (c) Express \(\dot{\phi}\) in terms of \(\ell\) and eliminate \(\dot{\phi}\) from the \(r\) equation. Now use the \(r\) equation to find the value \(r=r_{0}\) at which the first mass can move in a circular path. Interpret your answer in Newtonian terms. (d) Suppose the first mass is moving in this circular path and is given a small radial nudge. Write \(r(t)=r_{0}+\epsilon(t)\) and rewrite the \(r\) equation in terms of \(\epsilon(t)\) dropping all powers of \(\epsilon(t)\) higher than linear. Show that the circular path is stable and that \(r(t)\) oscillates sinusoidally about \(r_{\mathrm{o}}\) What is the frequency of its oscillations?

(a) Write down the Lagrangian \(\mathcal{L}\left(x_{1}, x_{2}, \dot{x}_{1}, \dot{x}_{2}\right)\) for two particles of equal masses, \(m_{1}=m_{2}=m,\) confined to the \(x\) axis and connected by a spring with potential energy \(U=\frac{1}{2} k x^{2} .\) [Here \(x\) is the extension of the spring, \(x=\left(x_{1}-x_{2}-l\right),\) where \(l\) is the spring's unstretched length, and I assume that mass 1 remains to the right of mass 2 at all times.] (b) Rewrite \(\mathcal{L}\) in terms of the new variables \(X=\frac{1}{2}\left(x_{1}+x_{2}\right)\) (the CM position) and \(x\) (the extension), and write down the two Lagrange equations for \(X\) and \(x\). (c) Solve for \(X(t)\) and \(x(t)\) and describe the motion.

A pendulum is made from a massless spring (force constant \(k\) and unstretched length \(l_{\mathrm{o}}\)) that is suspended at one end from a fixed pivot \(O\) and has a mass \(m\) attached to its other end. The spring can stretch and compress but cannot bend, and the whole system is confined to a single vertical plane. (a) Write down the Lagrangian for the pendulum, using as generalized coordinates the usual angle \(\phi\) and the length \(r\) of the spring. (b) Find the two Lagrange equations of the system and interpret them in terms of Newton's second law, as given in Equation (1.48). (c) The equations of part (b) cannot be solved analytically in general. However, they can be solved for small oscillations. Do this and describe the motion. [Hint: Let \(l\) denote the equilibrium length of the spring with the mass hanging from it and write \(r=l+\epsilon .\) "Small oscillations" involve only small values of \(\epsilon\) and \(\phi,\) so you can use the small-angle approximations and drop from your equations all terms that involve powers of \(\epsilon\) or \(\phi\) (or their derivatives) higher than the first power (also products of \(\epsilon\) and \(\phi\) or their derivatives). This dramatically simplifies and uncouples the equations.]

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