Problem 14
Question
A note on the piano has frequency \(F\). Suppose the maximum displacement at the center of the piano wire is given by \(s(0)\). Find constants a and \(\omega\) so that the equation \(s(t)=a \cos \omega t\) models this displacement. Graph s in the viewing window \([0,0.05]\) by \([-0.3,0.3]\). $$F=220 ; s(0)=0.06$$
Step-by-Step Solution
Verified Answer
The model is \( s(t) = 0.06 \cos(440\pi t) \).
1Step 1: Understand the parameters
We are given the frequency \( F = 220 \) and the maximum displacement \( s(0) = 0.06 \). The displacement is modeled by the equation \( s(t) = a \cos(\omega t) \). We need to find constants \( a \) and \( \omega \).
2Step 2: Find \( a \)
Since the maximum displacement occurs at \( t = 0 \) and is given by \( s(0) = 0.06 \), it follows that \( s(0) = a \cos(\omega \cdot 0) = a \cdot 1 = a \). Hence, \( a = 0.06 \).
3Step 3: Find \( \omega \) using the frequency
The frequency \( F = 220 \) corresponds to the number of cycles per second. The angular frequency \( \omega \) is related to \( F \) by the formula \( \omega = 2\pi F \). Thus, \( \omega = 2\pi \times 220 \).
4Step 4: Plug the constants into the equation
We now have \( a = 0.06 \) and \( \omega = 440\pi \). The equation modeling displacement is \( s(t) = 0.06 \cos(440\pi t) \).
5Step 5: Graph \( s(t) \)
Use a graphing tool or software to plot \( s(t) = 0.06 \cos(440\pi t) \) in the viewing window \([0,0.05]\) by \([-0.3,0.3]\). This will display the cosine wave over the specified interval. Note that \( s(t) \) oscillates with a period inverse to the frequency and amplitude \( 0.06 \).
Key Concepts
FrequencyAmplitudeAngular Frequency
Frequency
In trigonometric functions, frequency is essential for understanding the behavior of waves and vibrations, such as those produced by a piano note. Frequency, denoted as \( F \), measures how often a cycle repeats in one second. Its unit is hertz (Hz), and it relates to the sound's pitch. For our piano example, the frequency given is \( 220 \) Hz, implying the string vibrates 220 times per second.
This characteristic determines how fast the wave oscillates, making it critical for modeling sounds with precision.
This characteristic determines how fast the wave oscillates, making it critical for modeling sounds with precision.
- The formula \( s(t) = a \cos(\omega t) \) includes frequency through the angular frequency \( \omega \).
- Frequency affects the period of the wave, inversely related: \[ \text{Period} = \frac{1}{F} \]
Amplitude
Amplitude in trigonometric functions refers to the peak value of the wave, representing the maximum extent of a vibration or oscillation, measured from the position of equilibrium. In our piano exercise, the amplitude is the maximum displacement \( s(0) \) from rest, given as \( 0.06 \).
This value indicates the loudness of the sound produced, with higher amplitudes resulting in louder notes on a piano.
This value indicates the loudness of the sound produced, with higher amplitudes resulting in louder notes on a piano.
- The displacement formula \( s(t) = a \cos(\omega t) \) represents amplitude as the constant \( a \).
- Amplitude remains consistent unless energy input changes.
Angular Frequency
Angular frequency \( \omega \) is crucial for understanding how fast a wave form passes through its cycle. It relates linearly to the frequency \( F \), showing how the wave rotates in cycles per second when visualized in a circular context.
The formula connecting angular frequency to frequency is straightforward: \( \omega = 2\pi F \). Angular frequency is measurable in radians per second.
The formula connecting angular frequency to frequency is straightforward: \( \omega = 2\pi F \). Angular frequency is measurable in radians per second.
- In our exercise, we compute \( \omega = 2\pi \times 220 = 440\pi \).
- Angular frequency helps model the wave's progression in the cosine function \( s(t) = 0.06 \cos(440\pi t) \).
Other exercises in this chapter
Problem 13
What fraction of a complete revolution is each of the following angles? (a) \(180^{\circ}\) (b) \(40^{\circ}\) (c) \(1^{\circ}\)
View solution Problem 14
To show that sec(- \(x\) ) = sec \(x\) for all \(x\) in the domain, we begin by writing $$ \sec (-x)=\frac{1}{\cos (-x)} $$ and then use the fact that \(\cos (-
View solution Problem 14
Solve each right triangle. In each case, \(C=90^{\circ} .\) If angle information is given in degrees and minutes, give answers in the same way. If given in deci
View solution Problem 14
Complete the table with exact trigonometric function values. Do not use a calculator. $$\begin{array}{|c|c|c|c|c|c|c|} \hline \theta & \sin \theta & \cos \theta
View solution