Problem 41 - Algebra 2 and Trigonometry

Question

As stated in the Chapter Opener, sound can be thought of as vibrating air. Simple sounds can be modeled by a function h \((t)\) of the form $$\mathrm{h}(t)=\sin (2 \pi f t)$$ where the frequency \(f\) is in kilohertz \((\mathrm{kHz})\) and \(t\) is time. a. The frequency of "middle \(\mathrm{C}^{\prime \prime}\) is approximately 0.261 \(\mathrm{kHz}\). Graph two cycles of \(\mathrm{h}(t)\) for middle \(\mathrm{C} .\) b. The frequency of \(C_{3},\) or the \(C\) note that is one octave lower than middle \(C\) , is approximately 0.130 \(\mathrm{kHz}\) . On the same set of axes, graph two cycles of \(\mathrm{h}(t)\) for \(\mathrm{C}_{3}\). c. Based on the graphs from parts a and b, the periods of each function appear to be related in what way?

Step-by-Step Solution

Answer

Middle C's period is half that of C3, so it oscillates twice as fast as C3.

Step 1: Understanding the function

The given function is \( h(t) = \sin(2\pi f t) \). Here, \( f \) is the frequency in kilohertz and \( t \) is time in seconds. \( h(t) \) models the sound wave as a sine function.

Step 2: Calculating the period for middle C

Middle C has a frequency \( f = 0.261 \text{ kHz} \). The period \( T \) of the sine wave is given by \( T = \frac{1}{f} \). So, \( T_{\text{middle C}} = \frac{1}{0.261} \approx 3.83 \text{ milliseconds} \).

Step 3: Plotting two cycles for middle C

To plot two full cycles of the wave, calculate the time interval for two cycles as \( 2T = 2 \times 3.83 \approx 7.66 \) milliseconds. Use \( h(t) = \sin(2\pi \times 0.261 \times t) \) to plot the wave over this interval.

Step 4: Calculating the period for C3

For \( C_3 \), the frequency is \( f = 0.130 \text{ kHz} \). The period is \( T_{C_3} = \frac{1}{0.130} \approx 7.69 \text{ milliseconds} \).

Step 5: Plotting two cycles for C3

Two cycles of the \( C_3 \) wave will span \( 2T = 2 \times 7.69 \approx 15.38 \) milliseconds. Use the function \( h(t) = \sin(2\pi \times 0.130 \times t) \) to plot over this interval.

Step 6: Analyzing the relationship between periods

The period of middle C is approximately half the period of \( C_3 \). This means that middle C oscillates twice as fast as \( C_3 \). Each cycle of middle C fits two cycles within one cycle of \( C_3 \).

Key Concepts

Frequency

In the context of trigonometric functions, frequency refers to how often the sine wave completes a full cycle. Think of it as the wave's speed of vibration. In this problem, frequency is given in kilohertz (kHz), which means thousands of cycles per second. A higher frequency indicates that the wave oscillates more frequently within a given time.

• For the middle C note, the frequency is 0.261 kHz, meaning it completes 261 cycles in a second.
• The C₃ note, which is an octave lower, has a frequency of 0.130 kHz, completing only 130 cycles per second.

In general, the frequency directly influences the musical pitch we hear. Higher frequency sounds contain more cycles per second and are perceived as higher pitched, while lower frequency sounds are heard as lower pitched.

Sine Wave

A sine wave is one of the simplest forms of a waveform—it represents a smoothly oscillating curve. When we use a trigonometric function like \[ h(t) = \sin(2\pi f t) \]we're essentially describing this beautiful, periodic shape mathematically.

• The wave starts at zero, rises to a maximum point, falls back through zero, dips to a minimum, and then returns to zero to complete one cycle.
• In terms of sound, a sine wave represents a pure tone, which is a basic sound without harmonics.

The frequency of a sine wave determines how many such curves fit in one second, thus influencing the pitch. With sound waves modeled as sine functions, they provide us clear, regular oscillations—ideal for understanding concepts in both physics and music.

Period of a Function

The period of a function is the amount of time it takes for the wave to complete one full cycle. For sine functions related to sound waves, this is calculated as the reciprocal of the frequency:\[ T = \frac{1}{f} \]

• For middle C, with frequency 0.261 kHz, the period is about 3.83 milliseconds (ms).
• For C₃, with frequency 0.130 kHz, the period stretches to around 7.69 ms.

This duration plays a critical role in understanding the time it takes for an oscillation to return to its starting point. More frequent oscillations (higher pitch sounds) have shorter periods, while less frequent ones (lower pitch sounds) have longer periods. Observing the plotted cycles, we see how the shorter period of middle C allows it to oscillate within the longer period of C₃, perfectly illustrating their relationship.