Question

A massless spring has unstretched length \(l_{\mathrm{o}}\) and force constant \(k\). One end is now attached to the ceiling and a mass \(m\) is hung from the other. The equilibrium length of the spring is now \(l_{1}\). (a) Write down the condition that determines \(l_{1}\). Suppose now the spring is stretched a further distance \(x\) beyond its new equilibrium length. Show that the net force (spring plus gravity) on the mass is \(F=-k x\). That is, the net force obeys Hooke's law, when \(x\) is the distance from the equilibrium position \(-\) a very useful result, which lets us treat a mass on a vertical spring just as if it were horizontal. (b) Prove the same result by showing that the net potential energy (spring plus gravity) has the form \(U(x)=\) const \(+\frac{1}{2} k x^{2}\)

Short answer

The net force is \( F = -kx \), and the potential energy is \( U(x) = \text{const} + \frac{1}{2} k x^2 \).

Step-by-step solution

Understanding Equilibrium Length

When the mass is hung at equilibrium, the force exerted by the spring is equal to the gravitational force on the mass. This condition can be represented by the equation \( k(l_1 - l_0) = mg \), where \( l_1 \) is the equilibrium length of the spring, \( l_0 \) is the unstretched length, \( m \) is the mass, and \( g \) is the acceleration due to gravity.

Re-arranging for Equilibrium Length

By rearranging the equilibrium condition \( k(l_1 - l_0) = mg \), we solve for \( l_1 \) as follows: \( l_1 = \frac{mg}{k} + l_0 \). This gives the new length of the spring when a mass \( m \) is attached and hung at rest.

Analyzing Forces Beyond Equilibrium

When the spring is stretched by an additional distance \( x \), the force exerted by the spring is \( -k(l_1 + x - l_0) = -k(x + l_1 - l_0) \). Using the equilibrium condition \( l_1 - l_0 = \frac{mg}{k} \), the expression simplifies to \(-kx - mg + mg = -kx\). The gravitational force and the spring force offset, leaving only the term \(-kx\).

Reforming Potential Energy Expression

The potential energy from the spring when stretched an additional \( x \) is \( U_{spring} = \frac{1}{2}k(l_1 + x - l_0)^2 \). The gravitational potential energy is \( U_{gravity} = -mg(l_1 + x) \).

Simplifying Total Potential Energy

Substitute \( l_1 = \frac{mg}{k} + l_0 \) into the total potential energy expression. After expanding and simplifying, we find the net potential energy \( U(x) = - \text{const} - mg\left(\frac{mg}{k} + l_0 \right) + \frac{1}{2} k x^2 + mgx \), which simplifies to \( U(x) = \text{const} + \frac{1}{2} k x^2 \).

Conclusion on Net Force

The problem demonstrates that the force on the mass follows Hooke's Law when displaced by \( x \) from equilibrium: \( F = -kx \). This is valid as the net potential energy conforms to \( U(x) = \text{const} + \frac{1}{2} k x^2 \).

Key concepts

Equilibrium Condition

In the context of springs, the equilibrium condition is a critical concept. It describes the state where forces are balanced. When you hang a mass on a spring, the spring stretches until the force exerted by the spring balances the gravitational force pulling the mass downwards. This is expressed by Hooke's Law:

• Formula: \( k(l_1 - l_0) = mg \)
• Where \( k \) is the spring constant, \( l_1 \) is the equilibrium length of the spring, \( l_0 \) is the original length without load, \( m \) is the mass, and \( g \) is acceleration due to gravity.

At equilibrium, the upward force exerted by the spring equals the downward gravitational force on the mass, meaning the system is at rest. The equilibrium length can be calculated by rearranging the equation to find \( l_1 = \frac{mg}{k} + l_0 \). This provides the length of the spring once the mass is added, and is fundamental for further calculations involving spring motion.

Potential Energy

Understanding potential energy is essential when studying spring systems. The potential energy in a spring-mass system includes both the spring's potential energy and the gravitational potential energy of the mass.

• Spring Potential Energy: This refers to the energy stored when the spring is either compressed or stretched. The formula is given by \( U_{spring} = \frac{1}{2} k (l_1 + x - l_0)^2 \), where \( x \) is the distance stretched beyond the equilibrium length \( l_1 \).
• Gravitational Potential Energy: This is the energy due to the mass being at a height; calculated with \( U_{gravity} = -mg(l_1 + x) \).

The problem involves simplifying these energies into a single expression: \( U(x) = \text{const} + \frac{1}{2} k x^2 \). Here, the energy depends only on how much \( x \), the displacement from equilibrium, has occurred. This simplification shows that the energy behaves like a system only dependent on spring potential energy, a valuable realization when analyzing oscillatory motion.

Spring Force

The concept of spring force is crucial before diving into forces acting on a spring-mass system. Spring force essentially follows Hooke’s Law, where the force exerted by the spring is proportional to the distance it is stretched or compressed from its natural length. The formula depicting this relationship is:

• \( F = -k x \)
• This equation highlights that the spring force \( F \) is opposite in direction to the displacement \( x \) from equilibrium.

In simpler terms, if the spring is stretched downward, it pulls upward, and vice versa. By analyzing beyond equilibrium, if the spring is additionally stretched by \( x \), the net force including gravity becomes \( F = -k x \), reinforcing the validity of Hooke's Law even in vertical setups. Through this principle, it's verified that vertical motion behaves equivalently to horizontal motion under spring tension, simplifying the understanding of spring dynamics. This helps to predict the behavior of the system when displaced from its equilibrium position.