Problem 78 - Thomas Calculus

Question

Accelerations whose magnitudes are proportional to displacement Suppose that the position of a body moving along a coordinate line at time \(t\) is a. \(s=a \cos k t+b \sin k t\) b. \(s=a \cosh k t+b \sinh k t\) Show in both cases that the acceleration \(d^{2} s / d t^{2}\) is proportional to \(s\) but that in the first case it is directed toward the origin, whereas in the second case it is directed away from the origin.

Step-by-Step Solution

Answer

In case a, acceleration is \(-k^2 s\) towards the origin; in case b, it's \(k^2 s\) away from the origin.

Step 1: Differentiate Position Equation - Case a

Given the position function \(s = a \cos k t + b \sin k t\), we need to find its second derivative to determine the acceleration. First, differentiate \(s\) to find the velocity \(v = \frac{ds}{dt}\). The first derivative is \(v = \frac{ds}{dt} = -ak\sin(kt) + bk\cos(kt)\).

Step 2: Find Acceleration - Case a

Differentiate the velocity function to find the acceleration. \(a = \frac{d^2s}{dt^2} = \frac{d}{dt}(-ak\sin(kt) + bk\cos(kt)) = -ak^2\cos(kt) - bk^2\sin(kt)\).Therefore, the acceleration is \(a = -k^2(a \cos kt + b \sin kt) = -k^2 s\).

Step 3: Analyze Proportionality and Direction - Case a

Since \(a = -k^2 s\), the acceleration is proportional to the displacement by a factor of \(-k^2\). The negative sign indicates that the acceleration is directed towards the origin.

Step 4: Differentiate Position Equation - Case b

Given the position function \(s = a \cosh kt + b \sinh kt\), differentiate \(s\) once to find the velocity. \(v = \frac{ds}{dt} = ak\sinh(kt) + bk\cosh(kt)\).

Step 5: Find Acceleration - Case b

Differentiate the velocity function to find the acceleration for the second case. \(a = \frac{d^2s}{dt^2} = \frac{d}{dt}(ak\sinh(kt) + bk\cosh(kt)) = ak^2\cosh(kt) + bk^2\sinh(kt)\).Therefore, \(a = k^2(a \cosh kt + b \sinh kt) = k^2 s\).

Step 6: Analyze Proportionality and Direction - Case b

Since \(a = k^2 s\), the acceleration is proportional to the displacement by a factor of \(k^2\). The positive sign indicates that the acceleration is directed away from the origin.

Key Concepts

Harmonic Motion

Harmonic motion is a type of periodic motion where an object moves back and forth over the same path. It's characterized by a restoring force that is proportional to the displacement of the object. When studying differential equations, harmonic motion is often modeled with trigonometric functions, like cosine and sine. These functions help describe how the position of an object changes over time.

In simple harmonic motion, such as the one described by the function \(s = a \cos kt + b \sin kt\), the acceleration is found to be proportional to the position but acts in the opposite direction. This means the object accelerates toward the equilibrium position or the origin. It's like a weight on a spring that moves up and down, constantly pulled back toward the center.

Some key ideas in understanding harmonic motion include:

• The link between position and acceleration: They are proportional but opposite in direction.
• The role of constants \(a\), \(b\) and \(k\): These constants adjust the motion's amplitude, phase, and frequency.

Hyperbolic Functions

Hyperbolic functions, like \(\cosh\) (cosine hyperbolic) and \(\sinh\) (sine hyperbolic), resemble their trigonometric counterparts but have different properties and reference hyperbolas rather than circles. In our exercise, the position function \(s = a \cosh kt + b \sinh kt\) uses these hyperbolic functions to describe a type of motion distinct from harmonic motion.

In cases where hyperbolic functions are involved, the acceleration is still proportional to position. However, in this scenario, the acceleration is in the **same** direction as the displacement, indicating motion moving continually away from the origin. This type of behavior can be seen in systems where the motion is not bounded and continues to increase, like a cable suspended from two towers.

Hyperbolic functions provide insights into:

• Growth and decay patterns in unbound systems.
• The mathematical structure in equations involving exponential growth.

Second Derivative

The second derivative gives us a powerful insight into how a function's rate of change itself is changing. In motion, it tells us about acceleration, which is the rate of change of velocity over time. In the context of this exercise, knowing how to effectively compute the second derivative is crucial to understanding if the motion accelerates towards or away from a point.

To find the second derivative:

• First, derive the position function to get the velocity (the first derivative).
• Then, derive the velocity function to find acceleration (the second derivative).

The notion of proportionality between acceleration and position originates from the second derivative. In math terms, it is expressed simply as either \(a = -k^2 s\) or \(a = k^2 s\), where the signs influence the direction.

Key takeaways about second derivatives include:

• They reveal how forces change over time.
• They aid in understanding stability and behavior of systems in dynamic environments.