Problem 35 - Calculus Early Transcendentals

Question

Show that if \(c_{1}\) and \(c_{2}\) are any constants, the function $$x=x(t)=c_{1} \cos (\sqrt{\frac{k}{m}} t)+c_{2} \sin (\sqrt{\frac{k}{m}} t)$$ is a solution to the differential equation for the vibrating spring. (The corresponding motion of the spring is referred to as simple harmonic motion.)

Step-by-Step Solution

Answer

The function satisfies the simple harmonic motion differential equation.

Step 1: Identify the requirements

To verify whether the given function \(x(t) = c_{1} \cos \left( \sqrt{\frac{k}{m}} t \right) + c_{2} \sin \left( \sqrt{\frac{k}{m}} t \right)\) is a solution, we must check its compliance with the differential equation for simple harmonic motion, which is \(\frac{d^2 x}{dt^2} = - \frac{k}{m} x\).

Step 2: Differentiate the function once

First, we find \(\frac{dx}{dt}\):\[ \frac{dx}{dt} = \frac{d}{dt} \left( c_{1} \cos \left( \sqrt{\frac{k}{m}} t \right) + c_{2} \sin \left( \sqrt{\frac{k}{m}} t \right) \right) \]Using the chain rule, this becomes:\[ \frac{dx}{dt} = - c_{1} \sqrt{\frac{k}{m}} \sin \left( \sqrt{\frac{k}{m}} t \right) + c_{2} \sqrt{\frac{k}{m}} \cos \left( \sqrt{\frac{k}{m}} t \right) \]

Step 3: Differentiate the function a second time

Now differentiate \(\frac{dx}{dt}\) to find \(\frac{d^2 x}{dt^2}\):\[ \frac{d^2 x}{dt^2} = \frac{d}{dt} \left( - c_{1} \sqrt{\frac{k}{m}} \sin \left( \sqrt{\frac{k}{m}} t \right) + c_{2} \sqrt{\frac{k}{m}} \cos \left( \sqrt{\frac{k}{m}} t \right) \right) \]Applying the chain rule again:\[ \frac{d^2 x}{dt^2} = - c_{1} \frac{k}{m} \cos \left( \sqrt{\frac{k}{m}} t \right) - c_{2} \frac{k}{m} \sin \left( \sqrt{\frac{k}{m}} t \right) \]

Step 4: Compare with the differential equation

Substitute the expressions for \(x(t)\) and \(\frac{d^2 x}{dt^2}\) back into the differential equation:Given the differential equation is \(\frac{d^2 x}{dt^2} = - \frac{k}{m} x\), substitute:\[ - c_{1} \frac{k}{m} \cos \left( \sqrt{\frac{k}{m}} t \right) - c_{2} \frac{k}{m} \sin \left( \sqrt{\frac{k}{m}} t \right) = - \frac{k}{m} \left( c_{1} \cos \left( \sqrt{\frac{k}{m}} t \right) + c_{2} \sin \left( \sqrt{\frac{k}{m}} t \right) \right) \]Both sides are equal, confirming the function is a solution.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In the realm of physics, they are crucial for describing various phenomena such as motion, heat, and waves. One of the classic examples is the differential equation for simple harmonic motion, \[ \frac{d^2 x}{dt^2} = - \frac{k}{m} x, \]which describes the motion of a vibrating spring.

This equation means that the acceleration of the system (the second derivative of position with respect to time) is directly proportional but opposite in direction to its displacement. Understanding these relationships helps in modeling the expected behavior of systems under dynamic conditions, such as oscillations in springs. Solving these equations gives us the position function, which shows how the position of the object changes over time.

Chain Rule

The chain rule is a fundamental differentiation technique used when dealing with composite functions. In calculus, if you have a function within another function, the chain rule helps differentiate them effectively. For example, with our function \( x(t) = c_{1} \cos \left( \sqrt{\frac{k}{m}} t \right) + c_{2} \sin \left( \sqrt{\frac{k}{m}} t \right) \),differentiating directly would be difficult without the chain rule.

By applying the chain rule, we recognize:

• The derivative of \( \cos(u) \) is \(-\sin(u) \cdot \frac{du}{dt} \)
• The derivative of \( \sin(u) \) is \(\cos(u) \cdot \frac{du}{dt} \)

Thus, to differentiate functions inside trigonometric expressions, the chain rule is essential for navigating through complex relationships and obtaining the right solution.

Vibrating Spring

A vibrating spring showcases a type of mechanical wave known as simple harmonic motion, which resembles phenomena like a mass on a spring or a pendulum swinging back and forth. In this system, the spring force is the restoring force, pulling the object back towards its equilibrium position.

The mathematical model for this kind of motion generally involves sinusoidal functions, as seen here with \( x(t) = c_{1} \cos \left( \sqrt{\frac{k}{m}} t \right) + c_{2} \sin \left( \sqrt{\frac{k}{m}} t \right) \). This function characterizes how the object oscillates over time. By understanding these principles:

• One can determine how far and how fast something vibrates.
• Such analysis is key in designing systems like shock absorbers, building supports, and even musical instruments where vibrations must be controlled.

Simple harmonic motion, thus, is not only a fundamental concept in physics but also a foundation for various applications across engineering and technology.