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 \(\sqrt{\frac{3\lambda}{m}}\).