Problem 17 - Applied Mathematics

Question

Prove that the potential energy of a light elastic string of natural length \(I\) and modulus \(\lambda\) when stretched to a length of \((l+x)\) is \(\frac{1}{2} \lambda \frac{x^{2}}{l}\). Two points \(\mathrm{A}\) and \(\mathrm{B}\) are in a horizontal line at a distance \(3 l\) apart. A particle Pof mass \(m\) is joined to \(\mathrm{A}\) by a light inextensible string of length \(4 l\) and is joined to \(\mathrm{B}\) by a light elastic string of natural length \(/\) and modulus \(\lambda\). Initially \(\mathrm{P}\) is held at a point \(C\) in \(A B\) produced such that \(B C=l\), both strings being just taut, and is then released from rest. If \(\lambda=\frac{m g}{4}\) show that when \(\mathrm{AP}\) is vertical the speed of the particle is \(2 \sqrt{g l}\) and find the instantaneous value of the tension in the elastic string in this position. (J.M.B)

Step-by-Step Solution

Answer

The speed of the particle is \(2 \sqrt{gl}\). The tension in the elastic string is \(\frac{3mg}{4}\).

Step 1: Understand Potential Energy of Elastic String

The potential energy (PE) of an elastic string depends on its extension. The formula given is \ \[ PE = \frac{1}{2} \lambda \frac{x^2}{l} \], where \( \lambda \) is the modulus of the string, \( l \) is the natural length, and \( x \) is the extension from the natural length.

Step 2: Initial Configuration and Forces

Initially, the particle P is held at point C such that BC = l and both strings are taut, meaning there is no slack. Here, the inextensible string has length 4l, and the elastic string will have an initial length of AB + BC = 3l + l = 4l.

Step 3: Applying Energy Conservation

When P is at its lowest point (AP is vertical), we apply conservation of mechanical energy. Initial potential energy is only due to gravity and the elastic potential energy. We need to find the total mechanical energy at the starting position and at the vertical position to find the speed.

Step 4: Initial Potential Energy Calculation

Initially, the potential energy (PE) due to the height of P is mgh where h = l. Thus, \[ PE_{gravity} = mgl \] The elastic potential energy is zero since the system starts just taut with no extension beyond natural lengths.

Step 5: Final Potential Energy Calculation

At the lowest vertical position, the height of P changes. Since AP is now vertical, with length 4l, we need to calculate the kinetic and potential energy. Kinetic energy \(KE\) will be inset and Elastic potential energy will change as string 2 extends.

Step 6: Setting up and Solving using Energy Conservation

Initially, total energy is all potential energy from height: \[ E_{initial} = mgl \] At the lowest point (vertical) potential energy is \[ PE_{gravity,final} = 4mgl \] The kinetic energy at this point is present by speed v: \[ KE_{final} = \frac{1}{2} mv^2 \] By conservation of energy: \[ mgl = 4mgl + \frac{1}{2} mv^2 \], then, \[\frac{1}{2} mv^ 2 = (4mgl - mgl) = -3mgl \] and therefore: \[v = 2 \sqrt{gl} \] and Extension x of elastic string would be 4l-1 = 3l

Step 7: Calculating Tension in Elastic String

The tension T in elastic string is given by Hooke's law: \[ T = \lambda \times \frac{x}{l} \] Here, \( \lambda = \frac{mg}{4} \). For x = 3l, \[ T = (\frac{mg}{4}) \times \frac{3l}{l} = \frac{3mg}{4} \].

Key Concepts

Conservation of Energy

The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this exercise, we use this principle to understand the changes in energy as the particle moves.
Initially, the energy in the system is purely potential energy due to the particle's height and any stored elastic potential energy in the stretched string.
As the particle falls and moves, this potential energy is converted into kinetic energy, giving the particle speed. At the lowest position, any elastic potential energy, gravitational potential energy, and kinetic energy must equal the initial total energy of the system.

• Total initial energy = Potential Energy due to height
• Converted into Kinetic Energy and Elastic Potential Energy when the particle is in motion

Hooke's Law

Hooke's law describes the behavior of elastic materials, like the elastic string in our exercise. According to Hooke's law, the force exerted by an elastic string is directly proportional to the extension or compression it undergoes from its natural length.
Mathematically, it is given by:
\[ F = -k \times x \]
Here, \( F \) is the force, \( k \) is the spring constant (or modulus \( \lambda \) in the problem), and \( x \) is the extension/compression.
When the particle is released and the elastic string stretches, this force (or tension) is what tries to pull it back, converting the elastic potential energy into kinetic energy as per the law.

• Force increases with extension
• Directly related to stiffness (modulus \( \lambda \)) of the string

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy possessed by an object due to its position relative to a lower height. It depends on mass, gravity, and height.
The formula is \[ PE_{gravity} = mgh \]
In this exercise, initially, the particle has a certain GPE because of its vertical position relative to the ground. As it moves downward, this energy gets converted into kinetic energy and elastic potential energy because of the conservation of energy.

• Depends on mass \(m\), gravitational acceleration \(g\), and height \(h\).
• Converts to kinetic energy as object falls.

Kinetic Energy

Kinetic energy (KE) is the energy that an object has due to its motion. It depends on the mass of the object and the square of its velocity. The formula is \[ KE = \frac{1}{2} mv^2 \]
In the exercise, as the particle P falls, its gravitational potential energy is converted into kinetic energy. When the particle reaches its lowest point and moves fastest, this kinetic energy is at its maximum, satisfying the conservation of energy principle.

• Increases with the square of the velocity
• Directly proportional to the mass of the object