Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

\(\dagger\) 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

Expert verified
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

01

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: \(x-direction: T1x-T2x-T3x+T4x=0\) \(y-direction: T1y+T4y-T2y-T3y=0\) \(z-direction: T1z+T2z+T3z+T4z=m_{1g}\)
02

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: \(mg = \lambda\sqrt{8}\)
03

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: \(\omega_1= \omega_2= \sqrt{\frac{3\lambda}{m}}\)
04

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\({ }^{\dagger} \mathrm{A}\) double pendulum consists of a bob of mass \(m\) suspended under gravity from a fixed point by a light string of length \(a\), and a second bob, also of mass \(m\), suspended from the first by a further light string, also of length \(a\). Show that if the motion is near equilibrium and is confined to a vertical plane through the fixed point then the normal frequencies are \(\sqrt{(2 \pm \sqrt{2}) g / a}\). Describe the corresponding normal modes. Without doing further calculation, describe the normal modes when the motion is not confined to a vertical plane.

\({ }^{\dagger}\) A uniform rod \(A B\) of mass \(m\) and length \(2 a\) is suspended by two light inextensible strings \(P A\) and \(Q B\), each of length \(2 a\). The fixed points \(P\) and \(Q\) are at the same level, distance \(2 a\) apart; and in equilibrium, \(P A\) and \(Q B\) are vertical. The system performs small oscillations about its equilibrium configuration. Determine the normal frequencies and describe the normal modes.

\({ }^{\dagger}\) A particle \(A\) of mass \(m\) is suspended from a fixed point \(O\) by a light string of length \(3 a\). A second particle \(B\) of mass \(m\) is suspended from \(A\) by a second light string of length \(2 b\) and a third particle \(C\) of mass \(m\) is suspended from \(B\) by a third light string of length \(c\). The system can move in a vertical plane through \(O\). Let \(x, y, z\) denote the horizontal displacements of \(A, B, C\) from their equilibrium positions. Show that an approximate Lagrangian

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free