Problem 6 - Applied Mathematics

Question

A particle of mass \(2 m\) is attached to one end of an elastic string of modulus \(m g\) whose other end is fixed to a point \(P .\) The particle is dropped from P. It will first come to rest: (a) when the tension in the string = \(2 m g\), (b) when the kinetic energy is zero, (c) below the equilibrium position, (d) when the length of the string has doubled.

Step-by-Step Solution

Answer

The particle first comes to rest when the kinetic energy is zero.

Step 1 - Understand the Problem

A particle of mass 2m is attached to an elastic string with modulus mg. The particle is dropped from point P, and we need to determine when it will first come to rest among the given options.

Step 2 - Identify Initial Conditions

The initial position is at point P where the string is neither stretched nor compressed. The weight of the particle is given by the force due to gravity: \(2mg\).

Step 3 - Determine Elastic Force

The modulus of elasticity (or spring constant) of the string is \(mg\). When the string is stretched by a length x, the elastic force F is given by Hooke's Law: \(F = k \times x = mg \times x\).

Step 4 - Apply Energy Conservation

Determine the potential energy (gravitational and elastic) and the kinetic energy at various positions. Energy conservation tells us that the total mechanical energy remains constant.

Step 5 - Calculate Equilibrium Position

The equilibrium position is where the elastic force equals the gravitational force, i.e., \(mg \times x = 2mg\). Solve for x to find the equilibrium position: \[x = \frac{2mg}{mg} = 2\]

Step 6 - Consider Each Option

a) When the tension in the string = 2mg, the string is maximally stretched, implying the particle might have already come to rest. b) When kinetic energy is zero, it could be any point where the motion reverses or changes direction. c) Below the equilibrium position might indicate it's further stretched after equilibrium.d) When the length of the string has doubled: If the string's initial length is L, doubling it doesn’t typically align with normal spring/stretch equations.

Step 7 - Synthesize Findings

Given the maximum tension and zero kinetic energy conditions, and because the particle is dropped, point b) matches the condition where the kinetic energy is zero, indicating it comes to rest at the lowest point before moving back.

Key Concepts

Elastic Force

Elastic force is the force exerted by a material when it is deformed (such as being stretched or compressed). In this problem, the elastic string exerts a force opposing the stretching caused by the weight of the particle. This force follows Hooke's Law, which states that the force is directly proportional to the deformation. For our string, if it's stretched by a length x, the elastic force F can be described mathematically as: \( F = k \times x \) where k is the spring constant or modulus of elasticity. In this exercise, the modulus is given as mg, so the force becomes: \( F = mg \times x \)
Understanding elastic force helps us analyze the particle's movement and the point at which it might come to rest as it oscillates between tension and relaxation.

Equilibrium Position

The equilibrium position is where the forces acting on the particle balance out, and there's no net force causing movement. Here, the gravitational force pulling the particle down is balanced by the elastic force pulling it up.
For our case, the particle of mass 2m experiences a gravitational force:
\( F_{gravity} = 2mg \)
The elastic force opposing this is:
\( F_{elastic} = mg \times x \)
At equilibrium, these forces are equal:
\( mg \times x = 2mg \)
Simplifying this gives us:
\( x = 2 \)
So, the equilibrium position is when the string stretches by twice its initial length. Understanding this helps in analyzing at which points the particle changes direction or comes to rest.

Hooke's Law

Hooke's Law is critical in understanding the behavior of springs and elastic materials. It states that the force exerted by a spring is directly proportional to its extension or compression, up to its elastic limit. Mathematically, it can be expressed as:
\( F = k \times x \)
where F is the elastic force, k is the spring constant, and x is the distance the material is stretched or compressed. In our problem, the spring constant is given as mg, so the force becomes:
\( F = mg \times x \)
This law helps predict the force exerted for any given stretch in the string, crucial for analyzing how and where the particle comes to rest.

Kinetic Energy

Kinetic energy is the energy of motion. For a particle with mass m and velocity v, it is given by:
\( KE = \frac{1}{2} m v^2 \)
In our exercise, kinetic energy changes as the particle moves. When the particle is released, it accelerates under gravity, converting potential energy to kinetic energy. As the string stretches, energy is transferred to elastic potential energy. When kinetic energy becomes zero, the particle momentarily stops before reversing direction. This point is significant because it reflects where the forces balance temporarily, corresponding to the condition in the exercise where the kinetic energy is zero.