Question

(a) The force exerted by a one-dimensional spring, fixed at one end, is \(F=-k x,\) where \(x\) is the displacement of the other end from its equilibrium position. Assuming that this force is conservative (which it is) show that the corresponding potential energy is \(U=\frac{1}{2} k x^{2},\) if we choose \(U\) to be zero at the equilibrium position. (b) Suppose that this spring is hung vertically from the ceiling with a mass \(m\) suspended from the other end and constrained to move in the vertical direction only. Find the extension \(x_{\mathrm{o}}\) of the new equilibrium position with the suspended mass. Show that the total potential energy (spring plus gravity) has the same form \(\frac{1}{2} k y^{2}\) if we use the coordinate \(y\) equal to the displacement measured from the new equilibrium position at \(x=x_{\mathrm{o}}\) (and redefine our reference point so that \(U=0\) at \(y=0\) ).

Short answer

The potential energy is \( U = \frac{1}{2} k x^2 \); for a mass, \( x_o = \frac{mg}{k} \), and the total potential energy is \( \frac{1}{2} k y^2 \) defined from the new equilibrium.

Step-by-step solution

Determine the expression for the potential energy

The potential energy function, \( U(x) \), for a conservative force is found by integrating the negative of the force. Since the force exerted by the spring is \( F = -kx \), its corresponding potential energy \( U(x) \) is given by the integral \( U(x) = -\int F \, dx = \int kx \, dx \).

Integrate the force to find the potential energy

When performing the integration \( \int kx \, dx \), we obtain \( U(x) = \frac{1}{2} k x^2 + C \), where \( C \) is the constant of integration. To find the value of \( C \), we use the condition that \( U = 0 \) at the equilibrium position \( x = 0 \).

Apply boundary conditions to determine the constant

At the equilibrium position, \( x = 0 \), we set \( U(0) = 0 \). Substituting \( x = 0 \) into \( U(x) = \frac{1}{2} k x^2 + C \), we get \( 0 = \frac{1}{2} k (0)^2 + C = C \). Thus, \( C = 0 \), and the potential energy becomes \( U(x) = \frac{1}{2} k x^2 \).

Analyze the spring in vertical orientation

When the spring is hung vertically with a mass \( m \) suspended, the mass experiences the spring force and gravitational force. The gravitational force is \( mg \), where \( g \) is the acceleration due to gravity.

Find the new equilibrium position

At equilibrium, the spring force balances the gravitational force, so \( kx_o = mg \). From this, the extension at the new equilibrium position is \( x_o = \frac{mg}{k} \).

Redefine the potential energy in terms of displacement from new equilibrium

Define the coordinate \( y = x - x_o \), where \( x_o \) is the new equilibrium extension. The total potential energy, integrating both spring and gravity, is given by \( U = \frac{1}{2} k (y + x_o)^2 + mgy \).

Simplify the total potential energy expression

Simplify the expression obtained in Step 6: \( U = \frac{1}{2} k [(y + x_o)^2] + mg(y + x_o) - mgx_o \). By expanding and simplifying terms using \( kx_o = mg \), the potential energy simplifies to \( \frac{1}{2} k y^2 + U_{0} \), where the constant part is absorbed by redefining \( U = 0 \) at \( y = 0 \).

Key concepts

Harmonic Oscillator

A harmonic oscillator is a system where the force acting on an object is proportional to its displacement from a rest position and acts in the opposite direction. Imagine stretching a spring and letting go. It bounces back and forth around its equilibrium point. This is exactly what a harmonic oscillator does. The spring force, calculated as \( F = -kx \), acts to bring the object back to equilibrium. Here, \( k \) is the spring constant, which measures the stiffness of the spring. The larger \( k \), the stiffer the spring is.The harmonic motion is periodic, meaning it repeats itself in regular intervals. In ideal conditions — no energy loss through friction or air resistance — it continues indefinitely. The time it takes to make one complete cycle is known as the period. Harmonic oscillators apply to many systems, including pendulums, vibrating strings, and even electromagnetic circuits, where the concepts of cycles and restoring forces are crucial.

Conservative Forces

Conservative forces are those where the work done in moving an object from one point to another is independent of the path taken. Instead, it only depends on the starting and ending points.The best example is gravity and, in our case, the force exerted by a spring:

• For a spring, the work required to compress or extend the spring is returned as potential energy. This energy can be completely recovered, demonstrating that the force is conservative.

The potential energy associated with a conservative force can be defined mathematically. For a spring, the potential energy is calculated by integrating the force. For our spring, integrated the force \( F = -kx \) leads to the potential energy \( U = \frac{1}{2} k x^2 \). This shows that potential energy increases with the square of displacement, demonstrating the key principle of conservation of energy.

Vertical Spring-Mass System

In a vertical spring-mass system, the spring hangs from a ceiling with a mass attached at the bottom. Unlike the horizontal scenario, gravity becomes significant here. Both spring force and gravitational force act on the mass, affecting its motion.
The gravitational force on the mass is \( mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity. In equilibrium, the spring force balances the gravitational force making \( kx_o = mg \). Solving this gives the new equilibrium position extension \( x_o = \frac{mg}{k} \).In this system, the mass can move up and down while experiencing both forces. However, both forces are conservative, so the total mechanical energy in the system remains constant. When accounting for potential energy, both gravitational and spring energies are calculated together in the energy expressions to find conditions of movement and equilibrium.

Equilibrium Position

The equilibrium position in physics refers to the point where all forces acting on an object balance out to zero. In the context of a spring-mass system, it is the position of the mass when the net force from the spring and any other forces, like gravity, equals zero.
For a spring, the equilibrium position is initially defined as where the spring is neither compressed nor extended. In a vertical spring-mass system, it's where the spring force equals the gravitational force \( (kx_o = mg) \).If displaced from this point, the system attempts to return it through harmonic motion. Defining potential energy around this position (as \( U = 0 \) at equilibrium) helps in simplifying calculations, where potential energy becomes a function of displacement. Overall, recognizing and calculating displacements from such points are pivotal for analyzing the system's behavior in both static and dynamic scenarios.