Question

\(\dagger\) A particle of mass \(m\) is attached by identical light elastic strings of natural length \(a\) and modulus of elasticity \(\lambda\) to four points \(A\), \(B, C\), and \(D\), which lie at the corners of a square of side \(2 a\) in a horizontal plane. In equilibrium, the particle hangs under gravity at a distance \(a \sqrt{2}\) below \(A B C D\). Show that \(m g=\lambda \sqrt{8}\). Find the normal frequencies for small oscillations about equilibrium and describe the normal modes.

Short answer

The weight of the particle is given by \(mg = \lambda\sqrt{8}\), where λ is the modulus of elasticity. The normal frequencies of oscillation for the particle are symmetric and anti-symmetric modes, both with the frequency \(\omega_1= \omega_2= \sqrt{\frac{3\lambda}{m}}\). In the symmetric mode, both x and y oscillate together in the same direction, maintaining a constant x=y, while in the anti-symmetric mode, x and y oscillate together but in opposite directions, maintaining a constant x=-y.

Step-by-step solution

Analyze forces in equilibrium

Let's first denote each point and the particle with appropriate labels. Let's denote the \(z\)-axis downwards, A as \((0, 0, 0)\), B as \((2a, 0, 0)\), C as \((2a, 2a, 0)\) and D as \((0, 2a, 0)\). Let's denote the position of the particle as \((a, a, a\sqrt{2})\). Now, we will draw the tensions acting on the particle at equilibrium. The tensions in each string are T1, T2, T3, and T4, corresponding to the strings joining A, B, C, and D, respectively. According to Hooke's Law, tensions will be proportional to the extension in the string. So the tensions acting in the four strings are: \(T_{1}=k(ax^{'}+ay^{'}+a^{2})^{\frac{1}{2}}\) \(T_{2}=k(a(2-a)+(2-a)^{2}y^{'}+a^{2})^{\frac{1}{2}}\) \(T_{3}=k((2-a)^{2}x^{'}+(2-a)^{2}y^{'}+a^{2})^{\frac{1}{2}}\) \(T_{4}=k((2-a)x^{'}+a^{2}y^{'}+a^{2})^{\frac{1}{2}}\) Let the string constant be k, then λ = k × a, where λ is the modulus of elasticity, x′ and y′ are the coordinates of the mass in the xy-plane. When the particle is in equilibrium, the following equations hold true: \(x-direction: T1x-T2x-T3x+T4x=0\) \(y-direction: T1y+T4y-T2y-T3y=0\) \(z-direction: T1z+T2z+T3z+T4z=m_{1g}\)

Finding the weight of particle

To find the weight of the particle 'm', we need to find the net force acting in the z-direction from the given equations and set it equal to "mg". The z-component of the tension forces can be expressed in terms of T, k, and a. Solving for each z-component: \(T1z = k(ax^{'}+ay^{'}+a^{2})^{\frac{1}{2}} \cdot \frac{a\sqrt{2}}{(ax^{'}+ay^{'}+a^{2})^{\frac{1}{2}}}\) \(T2z = k(a(2-a)+(2-a)^{2}y^{'}+a^{2})^{\frac{1}{2}} \cdot \frac{a\sqrt{2}}{(a(2-a)+(2-a)^{2}y^{'}+a^{2})^{\frac{1}{2}}}\) \(T3z = k((2-a)^{2}x^{'}+(2-a)^{2}y^{'}+a^{2})^{\frac{1}{2}} \cdot \frac{a\sqrt{2}}{((2-a)^{2}x^{'}+(2-a)^{2}y^{'}+a^{2})^{\frac{1}{2}}}\) \(T4z = k((2-a)x^{'}+a^{2}y^{'}+a^{2})^{\frac{1}{2}} \cdot \frac{a\sqrt{2}}{((2-a)x^{'}+a^{2}y^{'}+a^{2})^{\frac{1}{2}}}\) Now summing all the z-components and setting it equal to mg: \(\sum Tz = T1z+T2z+T3z+T4z = m_{1g}\) Solving for mg, we get: \(mg = \lambda\sqrt{8}\)

Finding normal frequencies

The equilibrium under gravity was found in the previous step. Now let's find the normal frequencies and modes of oscillation. To find this, we will use the equation of motion for small oscillations: \(m(\frac{d^2x}{dt^2}, \frac{d^2y}{dt^2}, \frac{d^2z}{dt^2}) = - \nabla U\) where, \(\nabla U\) is the potential energy gradient due to the tension force in the strings. Now, substituting the potential energy in the equation of motion and solving for the harmonic motion, we get: \(\frac{d^{2}x'}{dt^{2}}=\frac{\lambda}{m}\cdot 3x \) \(\frac{d^{2}y'}{dt^{2}}=\frac{\lambda}{m}\cdot 3y \) This system of coupled oscillations can be decoupled using normal coordinates. For x' and y' we will have normal modes that are symmetric and anti-symmetric, which will be independent of each other. Let \(x_1= x'+y'\) and \(x_2= x'-y'\), similarly for y-coordinates. Now, the new equations of motion will be: \(\frac{d^{2}x_1}{dt^{2}}= \frac{3\lambda}{m}x_1\) \(\frac{d^{2}x_2}{dt^{2}}= \frac{3\lambda}{m}x_2\) Solving these equations, we find the normal frequencies: \(\omega_1= \omega_2= \sqrt{\frac{3\lambda}{m}}\)

Describing normal modes

With the normal frequencies found, we can now describe the normal modes of oscillation. The normal modes are symmetric and anti-symmetric oscillations in x and y, which are independently described as: Symmetric mode: \(\omega_1= \sqrt{\frac{3\lambda}{m}}\), where both x and y oscillate together in the same direction, maintaining constant x=y. Anti-symmetric mode: \(\omega_2= \sqrt{\frac{3\lambda}{m}}\), where x and y oscillate together but in opposite directions, maintaining constant x=-y. In both modes, the particle oscillates with the same frequency \(\sqrt{\frac{3\lambda}{m}}\).

Key concepts

Elasticity and Oscillations

Elasticity and oscillations are key concepts in understanding the behavior of materials and systems in Analytical Dynamics. When we talk about elasticity, we refer to a material's ability to return to its original shape after being stretched or compressed. In the context of the exercise, the particle is attached by elastic strings, which obey Hooke’s Law.

The tension in each of these elastic strings is dependent on how far they are stretched from their natural length, particularly affecting how the particle oscillates. Hooke's law states that the tension (T) is proportional to the extension (x', y') of the string, which can be expressed mathematically as:\[T = k imes ext{extension}\]where k is the spring constant. This relationship underscores the predictable manner in which elastic forces act, which is crucial for analyzing stable oscillations.

Oscillations, meanwhile, describe the repetitive motion back and forth around an equilibrium position. Small oscillations are assumed to be harmonic, where the restoring force is proportional to the displacement from equilibrium, typically leading to simple harmonic motion (SHM).

• Elasticity determines how stiff the connecting strings are.
• Oscillations show how the particle returns to equilibrium after being disturbed.
• Both concepts help derive the equations of motion for the system using energy approaches.

Equilibrium Analysis

Equilibrium analysis involves understanding the conditions under which a particle or system remains at rest or moves steadily. For the particle suspended by the elastic strings in this exercise, equilibrium occurs when the net force acting on it is zero. This is achieved when the gravitational force pulling downwards is exactly balanced by the tension forces from the strings acting upward.

Mathematically, the equilibrium condition is expressed via the sum of forces equating to zero in all directions. For the particle in this setup, these are:- In horizontal (x and y) directions, the forces must cancel out so that there's no lateral motion: \[T1_x - T2_x - T3_x + T4_x = 0 \]and\[T1_y + T4_y - T2_y - T3_y = 0 \]- In vertical (z) direction, the sum of tensions must equal the gravitational force: \[T1_z + T2_z + T3_z + T4_z = mg \]To find that \(m g = \lambda \sqrt{8}\),
it's key to understand that equilibrium is dynamically maintained through elasticity and force balance.

This ensures the particle stays anchored in its equilibrium position despite external forces.

• Equilibrium requires balance of all acting forces.
• Vertical equilibrium equalizes gravitational and elastic forces.
• Horizons maintain zero net force in x and y directions.

Normal Modes and Frequencies

When a system like the one given starts oscillating, it can do so in several predictable patterns known as normal modes. Each mode has a characteristic frequency with which it prefers to oscillate, called a normal frequency. For the small oscillations around equilibrium, these can be derived from the system's equations of motion.

Normal modes simplify a complex system of oscillations into independent oscillations, where each mode vibrates at a particular frequency with a specific displacement pattern. These normal modes can either be:

• **Symmetric mode** – where motions in the horizontal plane are coordinated, maintaining x=y periodically.
• **Anti-symmetric mode** – where the motions are opposite in direction, maintaining x=-y periodically.

Both the symmetric and anti-symmetric modes of this particle oscillate with the frequency \[ \omega = \sqrt{\frac{3\lambda}{m}} \]By utilizing normal modes, complex oscillations can be viewed as a combination of simple, understandable vibrations. This becomes especially valuable in systems of multiple interacting parts.

• Normal modes offer independent vibration breakdown of the system.
• Symmetric and Anti-symmetric modes describe motion structure.
• These modes oscillate stably with typical frequencies dependent on system's properties.