Problem 51 - Calculus Early Transcendentals

Question

Hooke's law is applicable to idealized (linear) springs that are not stretched or compressed too far. Consider a nonlinear spring whose restoring force is given by \(F(x)=16 x-0.1 x^{3},\) for \(|x| \leq 7\) a. Graph the restoring force and interpret it. b. How much work is done in stretching the spring from its equilibrium position \((x=0)\) to \(x=1.5 ?\) c. How much work is done in compressing the spring from its equilibrium position \((x=0)\) to \(x=-2 ?\)

Step-by-Step Solution

Answer

Answer: The work done in stretching the spring from its equilibrium position to x = 1.5 is approximately 16.875 Joules.

Step 1: a. Graph and Interpretation of Restoring Force

Since the problem doesn't require us to show the graphing process in detail, it is assumed that you can use a graphing tool to plot the function \(F(x) = 16x - 0.1x^3\). From the graph, you can observe that the restoring force follows a cubic function. As the spring is stretched or compressed, the restoring force increases or decreases respectively, but not linearly, due to the \(-0.1x^3\) term. This indicates that the spring behaves nonlinearly and has more complex behavior due to the cubic force term.

Step 2: b. Work Done in Stretching the Spring

To find the work done in stretching the spring from its equilibrium position \((x=0)\) to \(x=1.5\), we need to calculate the integral of the force function over the interval \([0,1.5]\). The work done, \(W\), can be represented as: $$ W = \int_{0}^{1.5} F(x) \, dx = \int_{0}^{1.5} (16x - 0.1x^3) \, dx $$ We can integrate the function and find the result as follows: $$ W = \left[8x^2 - 0.025x^4\right]_{0}^{1.5} $$ $$ W = (8(1.5)^2 - 0.025(1.5)^4) - (8(0)^2 - 0.025(0)^4) $$ $$ W \approx 16.875\,\text{J} $$ Therefore, the work done in stretching the spring from its equilibrium position to \(x=1.5\) is approximately \(16.875\) Joules.

Step 3: c. Work Done in Compressing the Spring

To find the work done in compressing the spring from its equilibrium position \((x=0)\) to \(x=-2\), we need to calculate the integral of the force function over the interval \([0,-2]\). The work done, \(W\), can be represented as: $$ W = \int_{0}^{-2} F(x) \, dx = \int_{0}^{-2} (16x - 0.1x^3) \, dx $$ Since the interval starts from \(0\) and goes to a negative number, the integral should be from \(-2\) to \(0\) with a negative sign to represent the compression of the spring. Thus, the integral should be written as: $$ W = -\int_{-2}^{0} (16x - 0.1x^3) \, dx $$ We can integrate the function and find the result as follows: $$ W = -\left[8x^2 - 0.025x^4\right]_{-2}^{0} $$ $$ W = -(8(0)^2 - 0.025(0)^4) - (8(-2)^2 - 0.025(-2)^4) = -(8 \cdot 4 - 0.025 \cdot 16) = -29.6\,\text{J} $$ In compressing the spring, we obtain a negative value for the work done, which represents the direction of the compression force. Therefore, approximately \(29.6\) Joules of work are done compressing the spring from its equilibrium position to \(x=-2\).

Key Concepts

Nonlinear Springs

In physics, most springs are considered to be linear, meaning they follow Hooke's Law, where the force needed to extend or compress the spring is proportional to the displacement. However, nonlinear springs do not adhere to this linear relationship. Instead, their behavior is more complex because the force depends on higher powers of displacement.

For the spring in our exercise, the restoring force is given by the equation:

• \( F(x) = 16x - 0.1x^3 \)

This equation shows that the spring's response is not simply a matter of multiplying the displacement by a constant. Instead, the presence of the \( -0.1x^3 \) term makes the force a cubic function, leading to nonlinear behavior. Nonlinear springs often result in more complex and interesting dynamics since the force varies in a way that is not directly proportional to displacement. This complexity is why nonlinear springs are found in applications requiring variable responses and custom force patterns.

Restoring Force

A restoring force is crucial in returning a deformed spring to its equilibrium position. In linear springs, this force is directly proportional to the displacement, following Hooke's Law. However, for the nonlinear spring in our example, the restoring force is more complex, as shown by the equation \( F(x) = 16x - 0.1x^3 \).

This function illustrates how the restoring force changes across different displacements:

• For small displacements, the force is relatively linear (approximately proportional to \( x \)).
• As the displacement increases, the \( -0.1x^3 \) term becomes more significant, introducing nonlinear effects.

At small deformations, the linear term (16x) has a dominant influence. However, at larger displacements, the cubic term (\( -0.1x^3 \)) modifies the force significantly, causing it to increase or decrease at a different rate. Therefore, this restoring force does not simply oppose the motion but offers a variable response based on how far the spring is extended or compressed.

Work Done in Stretching

When a spring is stretched, energy is required to extend it from its equilibrium position, resulting in work being done on the spring. For our nonlinear spring, calculating this work involves integrating the force function over the desired interval of stretching. The work done, \( W \), is computed as:

• \( W = \int_0^{1.5} (16x - 0.1x^3) \, dx \)

This integration yields the result:

• \( W \approx 16.875 \, \text{Joules} \)

This value represents the energy needed to stretch the spring from \( x = 0 \) to \( x = 1.5 \). Stretching a spring stores potential energy in it, which can be recovered when the spring attempts to return to its original position. Because of the nonlinear nature of the spring, the relationship between displacement and restoring force is more intricate than in linear springs, leading to a more complex calculation of work.

Work Done in Compressing

Just like stretching, compressing a spring requires work to be done, as energy must be expended to push the spring from its equilibrium position. For the provided nonlinear spring, the work done in compressing is found through integrating the restoring force over the compression interval. This is calculated as:

• \( W = -\int_{-2}^0 (16x - 0.1x^3) \, dx \)

Carrying out the integration gives the work done:

• \( W = -29.6 \, \text{Joules} \)

The negative sign indicates that the direction of force applied is opposite to the direction of the compressing motion. The value signifies the work necessary to compress the spring from \( x = 0 \) to \( x = -2 \). The calculation shows that regardless of whether a spring is compressed or stretched, energy must be transmitted into or out of the spring, aligning with notions of mechanical energy transfer. Nonlinear springs often demand these nuanced calculations due to their unique force-displacement relationships.