/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 The equation \(y=262(1+0.0593)^{... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The equation \(y=262(1+0.0593)^{x}\) models the frequency in hertz of various notes on the piano, with middle \(\mathrm{C}\) considered as note 0 . The average human ear can detect frequencies between 20 and 20,000 hertz If a piano keyboard were extended, the nighest and lowest notes audible to the average human ear would be how far above and below middle \(\mathrm{C}\) ? (it)

Short Answer

Expert verified
The highest audible note is approximately 96 notes above, and the lowest is approximately 63 notes below middle C.

Step by step solution

01

Analyzing the Equation

The given equation is \( y = 262(1 + 0.0593)^x \) where \( y \) is the frequency in hertz and \( x \) is the number of notes away from middle C. Middle C is defined as \( x = 0 \), giving a frequency of \( 262 \) Hz for middle C.
02

Define Limits of Human Hearing

The average human ear can detect frequencies between 20 Hz and 20,000 Hz. We need to solve for \( x \) when \( y = 20 \) Hz for the lowest frequency, and \( y = 20,000 \) Hz for the highest frequency.
03

Solving for Lowest Frequency (20 Hz)

For \( y = 20 \), substitute into the equation: \[ 20 = 262(1 + 0.0593)^x \] Divide both sides by 262 to isolate the exponent: \[ \frac{20}{262} = (1 + 0.0593)^x \] Calculates to approximately \( 0.0763 = (1.0593)^x \). Use logs to solve for \( x \): \[ x = \frac{\log(0.0763)}{\log(1.0593)} \approx -63.37 \] This means the lowest note corresponds to \( x \approx -63 \) below middle C.
04

Solving for Highest Frequency (20,000 Hz)

For \( y = 20,000 \), substitute into the equation:\[ 20,000 = 262(1 + 0.0593)^x \] Divide both sides by 262:\[ \frac{20,000}{262} = (1.0593)^x \] Calculates to approximately \( 76.34 = (1.0593)^x \).Use logs to solve for \( x \): \[ x = \frac{\log(76.34)}{\log(1.0593)} \approx 96.31 \] This means the highest note corresponds to \( x \approx 96 \) above middle C.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frequency of Sound
Sound is a wave that travels through the air, and frequency refers to how many waves pass a point in one second. This is measured in hertz (Hz). Imagine the waves are like cars passing by on a road; the more cars that pass by per second, the higher the frequency. When the frequency changes, the pitch of the sound also changes. High frequencies lead to higher pitched sounds, while low frequencies result in lower pitched sounds. In the context of musical notes, each note on a piano has a specific frequency.

For example, middle C, a common reference point, has a frequency of 262 Hz. This means 262 sound waves reach your ear each second when middle C is played. The equation provided, \( y = 262(1 + 0.0593)^x \), is used to calculate the frequencies of other piano notes relative to middle C. Here, \( x \) is the number of notes away from middle C. When \( x \) is zero, we get the 262 Hz frequency of middle C.
Human Hearing Range
Humans can hear a wide range of frequencies. This range is typically said to be from 20 Hz to 20,000 Hz, although it can vary from person to person. Think of this range as the limits of what we can experience through sound. Sounds below 20 Hz are considered infrasonic, while those above 20,000 Hz are ultrasonic, both usually beyond the normal range of human hearing.

In our math problem, we use this hearing range to determine which notes on an extended piano are audible. We calculate this by plugging the boundaries of our hearing range, 20 Hz and 20,000 Hz, into the exponential equation to find how far these notes are from middle C. By doing so, we determine the notes the average person can hear on a hypothetical piano that extends beyond the standard layout.
Mathematical Modeling
Mathematical modeling uses mathematical equations to represent real-world phenomena. In our example, we use the equation \( y = 262(1 + 0.0593)^x \) to model how piano notes map to frequencies. This model lets us relate the mathematical concept of exponential growth to a practical situation like music.

Why exponential? Because each successive note on a piano is a fixed percentage higher in frequency than the last one, which is a key characteristic of exponential functions. By substituting different values for \( x \), you can calculate the frequency of any note on an extended piano.

This kind of modeling is valuable because it helps to make predictions and understand relationships in an easy-to-calculate way. It connects abstract math to our daily experiences and can explain things like music, sound waves, and their frequencies in a straightforward yet powerful manner.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

APPLICATION In the course of a mammal's lifetime, \(i\) heart beats about 800 million times, regardless of the mammal's size or weight. (This excludes humans.) a. An elephant's heart beats approximately 25 times a minute. How many years would you expect an elephant to live? Use scientific notation to calculate your answer. (a) b. A pygmy shrew's heart beats approximately 1150 times a minute. How many years would you expect a pygmy shrew to live? c. If this relationship were true for humans, how many years would you expect a human being with a heart rate of 60 bpm to live?

A light-year is the distance light can travel in one year. This distance is approximately 9,460 billion kilometers. Dead skin cells are one The Milky Way galaxy is estimated to be about components of dust. 100,000 light-years in diameter. a. Write both distances in scientific notation. b. Find the diameter of the Milky Way in kilometers. Use scientific notation. c. Scientists estimate the diameter of Earth is greater than \(1.27 \times 10^{4} \mathrm{~km}\). How many times larger is the diameter of the Milky Way?

There are approximately \(5.58 \times 10^{21}\) atoms in a gram of silver. How many atoms are there in 3 kilograms of silver? Express your answer in scientific notation. (a)

Use the equation \(y=47(1-0.12)^{x}\) to answer each question. a. Does this equation model an increasing or decreasing pattern? (ii) b. What is the rate of increase or decrease? c. What is the \(y\)-value when \(x\) is 13 ? d. What happens to the \(y\)-values as the \(x\)-values get very large?

Use the distributive property and the properties of exponents to write an equivalent expression without parentheses. Use your calculator to check your answers, as you did in Exercise \(1 .\) a. \(x\left(x^{3}+x^{4}\right)\) b. \(\left(-2 x^{2}\right)\left(x^{2}+x^{4}\right)\) c. \(2.5 x^{4}\left(6.8 x^{3}+3.4 x^{4}\right)\) Write an equivalent expression in the form \(a \cdot b^{n}\). (T1)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.