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 \begin{equation}\begin{aligned} \text { a. } & s=a \cos k t+b \sin k t \\\ \text { b. } & s=a \cosh k t+b \sinh k t \end{aligned}\end{equation} 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 \( -k^2s \) points toward the origin; in (b), \( k^2s \) points away.

Step 1: Differentiate Once for Velocity (Case a)

Given the position function for case (a) is \( s = a \cos kt + b \sin kt \). We start by differentiating with respect to \( t \) to find the velocity: \\[ v = \frac{ds}{dt} = \frac{d}{dt}(a \cos kt + b \sin kt) = -ak \sin kt + bk \cos kt \].

Step 2: Differentiate Again for Acceleration (Case a)

Differentiate the velocity function to find acceleration: \\[ \frac{d^2 s}{dt^2} = \frac{d}{dt}(-ak \sin kt + bk \cos kt) = -ak^2 \cos kt - bk^2 \sin kt \]. \Notice that this can be rewritten using the original position function: \\[ \frac{d^2 s}{dt^2} = -k^2(a \cos kt + b \sin kt) = -k^2s \]. \This shows that the acceleration is proportional to \( s \) and is directed toward the origin because of the negative sign.

Step 3: Differentiate Once for Velocity (Case b)

For case (b), the position function is \( s = a \cosh kt + b \sinh kt \). Differentiate it with respect to \( t \) to find the velocity: \\[ v = \frac{ds}{dt} = \frac{d}{dt}(a \cosh kt + b \sinh kt) = ak \sinh kt + bk \cosh kt \].

Step 4: Differentiate Again for Acceleration (Case b)

Differentiate the velocity to find the acceleration: \\[ \frac{d^2 s}{dt^2} = \frac{d}{dt}(ak \sinh kt + bk \cosh kt) = ak^2 \cosh kt + bk^2 \sinh kt \]. \This can also be rewritten using the original position function: \\[ \frac{d^2 s}{dt^2} = k^2(a \cosh kt + b \sinh kt) = k^2s \]. \Here, acceleration is proportional to \( s \) and directed away from the origin because the proportionality constant \( k^2 \) is positive.

Key Concepts

Position Function

A position function describes how the position of an object changes over time. In other words, it tells us where the object is located at any given time while it's moving along a path. For this exercise, we have two types of position functions that involve trigonometric and hyperbolic functions.

In case (a), the position function is given by:

• \( s = a \cos(kt) + b \sin(kt) \)

This function uses cosine and sine, which are periodic functions capturing oscillating motion, usually seen in systems like pendulums or springs. Thus, this position is about periodic motion around the origin.

In case (b), the position function is:

• \( s = a \cosh(kt) + b \sinh(kt) \)

Here, cosine hyperbolic (cosh) and sine hyperbolic (sinh) functions describe growth or decay processes rather than periodic motion, reflecting the position for a scenario that involves exponential change over time.

Velocity Function

The velocity function arises from the derivative of the position function, which describes the rate at which the position of an object changes over time. Essentially, velocity tells us how fast and in what direction an object is moving at any particular point.

For case (a), the velocity function, derived from the position function \( s = a \cos(kt) + b \sin(kt) \), is:

• \( v = \frac{ds}{dt} = -ak \sin(kt) + bk \cos(kt) \)

This equation tells us that the velocity is also periodic as it involves sine and cosine terms.

In case (b), with the position \( s = a \cosh(kt) + b \sinh(kt) \), we derive:

• \( v = \frac{ds}{dt} = ak \sinh(kt) + bk \cosh(kt) \)

This velocity indicates a non-periodic but exponentially changing speed and direction.

Acceleration

The acceleration is found by differentiating the velocity function, which gives us the rate of change of velocity or how the speed of the object changes over time.

In case (a), the acceleration from a velocity of \( v = -ak \sin(kt) + bk \cos(kt) \) is:

• \( \frac{d^2s}{dt^2} = -ak^2 \cos(kt) - bk^2 \sin(kt) \)

Notice that this can be rewritten using the original position function: \( \frac{d^2s}{dt^2} = -k^2s \). The negative sign indicates that acceleration is proportional to the position, but directed towards the origin.

In case (b), with velocity \( v = ak \sinh(kt) + bk \cosh(kt) \), we get:

• \( \frac{d^2s}{dt^2} = ak^2 \cosh(kt) + bk^2 \sinh(kt) \)

Rewriting gives \( \frac{d^2s}{dt^2} = k^2s \). Here, acceleration is directed away, showing that the force increases with distance.

Hyperbolic Functions

Hyperbolic functions, such as hyperbolic sine (\( \sinh \)) and hyperbolic cosine (\( \cosh \)), play a crucial role in many areas of mathematics and physics, including when solving differential equations related to this exercise. Unlike their trigonometric counterparts that describe circular motion, hyperbolic functions describe hyperbolic or exponential growth paths.

They are defined as:

• Hyperbolic sine: \( \sinh(x) = \frac{e^x - e^{-x}}{2} \)
• Hyperbolic cosine: \( \cosh(x) = \frac{e^x + e^{-x}}{2} \)

These functions are important because they allow for solutions in problems involving exponential growth or decay. In the context of case (b), their role is critical in describing non-periodic motion paths where acceleration and changes in velocity are consistently directed away or towards in a linear manner.