Problem 254 - Calculus Volume 1

Question

A mass hanging from a vertical spring is in simple harmonic motion as given by the following position function, where \(t\) is measured in seconds and \(s\) is in inches: \(s(t)=-3 \cos \left(\pi t+\frac{\pi}{4}\right)\). a. Determine the position of the spring at \(t=1.5 \mathrm{~s}\). b. Find the velocity of the spring at \(t=1.5 \mathrm{~s}\).

Step-by-Step Solution

Answer

The position is \(-\frac{3\sqrt{2}}{2}\) inches; the velocity is \(-\frac{3\pi\sqrt{2}}{2}\) inches per second.

Step 1: Identify the position function

The given position function of the mass is \(s(t) = -3 \cos \left(\pi t + \frac{\pi}{4}\right)\). Our task is to find the position when \(t = 1.5\) seconds.

Step 2: Calculate the position at t = 1.5 seconds

Substitute \(t = 1.5\) into the position function: \[ s(1.5) = -3 \cos \left( \pi \cdot 1.5 + \frac{\pi}{4} \right) \]Calculate the angle inside the cosine: \[ s(1.5) = -3 \cos \left(1.5\pi + \frac{\pi}{4} \right) = -3 \cos \left( \frac{7\pi}{4} \right) \]Since \(\cos \left( \frac{7\pi}{4} \right) = \frac{\sqrt{2}}{2} \) (using trigonometric identities), we find: \[ s(1.5) = -3 \times \frac{\sqrt{2}}{2} = -\frac{3\sqrt{2}}{2} \text{ inches.} \]

Step 3: Determine the velocity function

The velocity, \(v(t)\), is the derivative of the position function \(s(t)\). Given \(s(t) = -3 \cos \left(\pi t + \frac{\pi}{4}\right)\), let us differentiate it with respect to \(t\).\[ v(t) = \frac{d}{dt}\left[-3 \cos \left(\pi t + \frac{\pi}{4}\right)\right] \]Using the chain rule: \[ v(t) = -3 \times (-\sin(\pi t + \frac{\pi}{4})) \times \pi \]\Thus, \[ v(t) = 3\pi \sin(\pi t + \frac{\pi}{4}) \]

Step 4: Calculate the velocity at t = 1.5 seconds

Substitute \(t = 1.5\) into the velocity function:\[ v(1.5) = 3\pi \sin \left( \pi \cdot 1.5 + \frac{\pi}{4} \right) \]Calculate the angle inside the sine:\[ v(1.5) = 3\pi \sin \left( \frac{7\pi}{4} \right) \] Since \(\sin \left( \frac{7\pi}{4} \right) = -\frac{\sqrt{2}}{2} \) (using trigonometric identities), the velocity is:\[ v(1.5) = 3\pi \times -\frac{\sqrt{2}}{2} = -\frac{3\pi\sqrt{2}}{2} \text{ inches per second.} \]

Key Concepts

Position Function

The position function in simple harmonic motion tells us how far a mass is from its equilibrium point at any given time. In this case, the equation is given by \[ s(t) = -3 \cos \left(\pi t + \frac{\pi}{4}\right) \]This is a cosine function, which is often used in simple harmonic motion to describe cyclical phenomena such as oscillating springs. Here, the minus sign indicates the direction of the initial displacement. The amplitude, which is 3, represents the maximum displacement from the equilibrium.
To find the position of the mass at a specific time, like at \(t = 1.5\) seconds, you plug the value of \(t\) into the position function. This involves the following steps:

• Calculate the angle inside the cosine: \(1.5\pi + \frac{\pi}{4} = \frac{7\pi}{4}\).
• Use trigonometric identities: \(\cos \left( \frac{7\pi}{4} \right) = \frac{\sqrt{2}}{2}\).
• Substitute back to find \(s(1.5) = -3 \times \frac{\sqrt{2}}{2} = -\frac{3\sqrt{2}}{2}\) inches.

This result tells us where the mass is positioned at exactly \(1.5\) seconds.

Velocity Function

Understanding how fast the mass is moving is crucial, and that's where the velocity function comes in. The velocity function is simply the derivative of the position function. Let's break down the steps:

• Given the position function \(s(t) = -3 \cos \left(\pi t + \frac{\pi}{4}\right)\).
• Differentiate this function with respect to \(t\) using the chain rule: \[ v(t) = \frac{d}{dt} \left[ -3 \cos \left(\pi t + \frac{\pi}{4}\right) \right] = 3\pi \sin(\pi t + \frac{\pi}{4}) \]

This derivative tells us how the position changes over time, describing the speed of the oscillation.
To find the specific velocity at \(t = 1.5\) seconds, you follow the steps:

• Calculate the angle inside the sine: \(1.5\pi + \frac{\pi}{4} = \frac{7\pi}{4}\).
• Use trigonometric identities: \(\sin \left( \frac{7\pi}{4} \right) = -\frac{\sqrt{2}}{2}\).
• Substitute back to find \(v(1.5) = 3\pi \times -\frac{\sqrt{2}}{2} = -\frac{3\pi\sqrt{2}}{2}\) inches per second.

This tells us the speed and direction of the mass at \(1.5\) seconds.

Trigonometric Identities

Trigonometric identities are essential tools in solving problems involving harmonic motion. They help simplify and transform expressions to find exact values for functions like sine and cosine.
For our exercise, two key trigonometric identities were used:

• \( \cos \left( \frac{7\pi}{4} \right) = \frac{\sqrt{2}}{2} \)
• \( \sin \left( \frac{7\pi}{4} \right) = -\frac{\sqrt{2}}{2} \)

These identities are derived from the unit circle, which is a fundamental concept in trigonometry. It helps us visualize the cyclic nature of these functions.
Understanding how to employ these identities allows you to solve complex trigonometric equations and simplify calculations, especially in periodic motion contexts like springs and pendulums. When you see angles like \(\frac{7\pi}{4}\), you'll often resort to these identities in problems related to oscillating systems.