Problem 69 - Precalculus: Functions and Graphs

Question

Motion of a mass If a mass that is attached to a spring is raised \(y_{0}\) feet and released with an initial vertical velocity of \(v_{0}\) ft/sec, then the subsequent position \(y\) of the mass is given by $$ y=y_{0} \cos \omega t+\frac{v_{0}}{\omega} \sin \omega t $$where \(t\) is time in seconds and \(\omega\) is a positive constant. (a) If \(\omega=1, y_{0}=2 \mathrm{ft},\) and \(v_{0}=3 \mathrm{ft} / \mathrm{sec},\) express \(y\) in the form \(A \cos (B t-C),\) and find the amplitude and period of the resulting motion. (b) Determine the times when \(y=0\) - that is, the times when the mass passes through the equilibrium position.

Step-by-Step Solution

Answer

(a) \( A = \sqrt{13} \), period \( T = 2\pi \). (b) \( t = \tan^{-1}(-\frac{2}{3}) + n\pi \).

Step 1: Identify values and expression for part (a)

Given values are \( y_0 = 2 \, \text{ft} \), \( v_0 = 3 \, \text{ft/sec} \), and \( \omega = 1 \). The expression for the motion is \( y = y_0 \cos \omega t + \frac{v_0}{\omega} \sin \omega t \). Substitute these into the expression: \( y = 2 \cos t + 3 \sin t \).

Step 2: Transform expression to standard form

The expression \( y = 2 \cos t + 3 \sin t \) needs to be transformed to the form \( A \cos (Bt - C) \). This can be done using the identity \( A \cos(x - \varphi) = A \cos \varphi \cdot \cos x + A \sin \varphi \cdot \sin x \). Compare this with our equation and identify that \( A \cos \varphi = 2 \) and \( A \sin \varphi = 3 \).

Step 3: Calculate amplitude \(A\)

Amplitude \( A \) is given by \( A = \sqrt{(A\cos \varphi)^2 + (A\sin \varphi)^2} = \sqrt{2^2 + 3^2} = \sqrt{13} \).

Step 4: Determine phase shift \(\varphi\)

Using \( \tan \varphi = \frac{3}{2} \), find \( \varphi \) such that \( \cos \varphi = \frac{2}{\sqrt{13}} \) and \( \sin \varphi = \frac{3}{\sqrt{13}} \). Thus, \( \varphi \) is the angle whose cosine is \(\frac{2}{\sqrt{13}}\) and sine is \(\frac{3}{\sqrt{13}}\).

Step 5: Identify the period

The period \( T \) of the function \( y = A \cos(t - C) \) is given by \( T = \frac{2\pi}{B} \). Here, \( B = 1 \), so \( T = 2\pi \).

Step 6: Solve for times when \(y = 0\)

The equation to solve is \( y = 2 \cos t + 3 \sin t = 0 \). Rearranging gives \( \tan t = -\frac{2}{3} \). Solving for \( t \), we get \( t = \tan^{-1}(-\frac{2}{3}) + n\pi \), where \( n \) is an integer.

Key Concepts

Trigonometric Identities

Trigonometric identities are incredibly useful in solving problems involving harmonic motion. These identities allow us to transform complex expressions into simpler forms. One common transformation is using the identity \(A \cos(x - \varphi) = A \cos \varphi \cdot \cos x + A \sin \varphi \cdot \sin x\). This identity is particularly helpful when you need to express a function as a single cosine wave.
For the given problem, the original motion equation is \(y = 2 \cos t + 3 \sin t\). To express it in the form of a single trigonometric function \(A \cos(Bt - C)\), we need to recognize \(A \cos \varphi = 2\) and \(A \sin \varphi = 3\). By using the Pythagorean identity, the expression can be rewritten to help calculate the amplitude, and identify the phase shift and period later on.

Amplitude

Amplitude is an important characteristic of wave motion. It represents the maximum displacement from the equilibrium position. For the function \(y = A \cos(Bt - C)\), the amplitude is simply \(|A|\).
Using the problem’s equation \(y = 2 \cos t + 3 \sin t\), we calculated the amplitude using the formula \(A = \sqrt{(A \cos \varphi)^2 + (A \sin \varphi)^2}\). Substituting the coefficients, we find \(A = \sqrt{2^2 + 3^2} = \sqrt{13}\). This value of \(A\) tells us how far the mass moves from its resting, or equilibrium, position during its motion.

Phase Shift

Phase shift indicates how much the wave is shifted horizontally from the usual position. To find the phase shift, we use the identity concerning cosine and sine components.
For our function \(y = 2 \cos t + 3 \sin t\), we recognize the transformations required to match it to \(A \cos(x - \varphi)\) form, where \(\tan \varphi = \frac{3}{2}\). Solving for \(\varphi\), we find that \(\cos \varphi = \frac{2}{\sqrt{13}}\) and \(\sin \varphi = \frac{3}{\sqrt{13}}\), showing the phase shift as the angle whose tangent is \(\frac{3}{2}\). This angle effectively tells you how much to "slide" the wave along the horizontal axis.

Period of a Function

The period of a function describes how long it takes for the wave to complete one full cycle. It is critical for understanding the timing of repetitive motion.
For harmonic motion described by \(y = A \cos(Bt - C)\), the formula for the period \(T\) is given by \(T = \frac{2\pi}{B}\). In our example with \(B = 1\), this simplifies the period to \(T = 2\pi\). This tells us that the mass on the spring takes \(2\pi\) seconds to return to a similar position and pattern as it started. Knowing the period is essential for predicting the timing of the wave's crest and trough times and can help in determining stability in systems like this spring mass.