Problem 51 - Physics for Scientists and Engineers with Modern Physics

Question

A certain audio compact disc (CD) contains 783.216 megabytes of digital information. Each byte consists of exactly 8 bits. When played, a CD player reads the CD's digital information at a constant rate of 1.4 megabits per second. How many minutes does it take the player to read the entire CD?

Step-by-Step Solution

Answer

It takes approximately 74.59 minutes for the player to read the entire CD.

Step 1: Determine Total Bits on CD

Begin by calculating the total number of bits on the CD. A byte consists of 8 bits, and the CD contains 783.216 megabytes of information. Thus, the total bits can be calculated as follows: \[ 783.216 \text{ megabytes} \times 10^6 \text{ bytes/megabyte} \times 8 \text{ bits/byte} = 6265728000 \text{ bits}. \]

Step 2: Convert Player's Reading Rate to Bits per Second

The player's reading rate is given in megabits per second (Mb/s). Convert this rate to bits per second by multiplying by \(10^6\): \[ 1.4 \text{ Mb/s} \times 10^6 \text{ bits/Mb} = 1400000 \text{ bits/second}. \]

Step 3: Calculate Total Time in Seconds

Use the total number of bits and the reading rate to determine the total time in seconds needed for the player to read the entire CD. Use the formula \( t = \frac{\text{Total bits}}{\text{Reading rate}} \): \[ t = \frac{6265728000 \text{ bits}}{1400000 \text{ bits/second}} \approx 4475.52 \text{ seconds}. \]

Step 4: Convert Total Time to Minutes

Convert the total time from seconds to minutes by dividing by 60: \[ 4475.52 \text{ seconds} \div 60 \approx 74.59 \text{ minutes}. \]

Key Concepts

Information Storage

Information storage in the context of digital devices like compact discs (CDs) is all about how data is saved. Data on a CD is stored in megabytes (MB), where each MB consists of a million bytes. A byte is a unit of digital information that characteristically holds eight bits. Bits are the smallest units of data in computing, represented as binary digits: either 0 or 1. When we talk about the storage capacity of a CD, we refer to the total amount of megabytes or bytes it can hold. In this exercise, the CD holds 783.216 megabytes of information. To understand the scale, this means the CD contains approximately 6265728000 bits, as each megabyte consists of a million bytes, with each byte comprising eight bits.

Data Reading Rate

The data reading rate for devices like a CD player is crucial because it indicates how quickly data can be read from the storage medium. The reading rate in this scenario is given as 1.4 megabits per second. To make calculations easier, this is often converted to bits per second. Since one megabit is equivalent to one million bits, multiplying 1.4 by 1,000,000 gives a reading rate of 1,400,000 bits per second. This conversion is vital in determining how long it will take a CD player to read the entire data stored on a CD. Knowing the exact reading rate allows us to estimate the time needed to access and utilize the information stored digitally.

Time Conversion

Time conversion is necessary when working with different units of measure in calculations, especially when determining the duration for processes like data reading. In this problem, the calculated reading time is first determined in seconds. Using a basic conversion where one minute equals 60 seconds, you can easily find out how many minutes it will take for a task by dividing the total seconds by 60. For instance, if the total time is 4,475.52 seconds, it converts to approximately 74.59 minutes when divided by 60. This kind of conversion helps us understand results in a more relatable time frame, which is essential for practical applications.

Digital Information

Digital information refers to data that is stored and processed using electronic or digital means. On a CD, digital information is encoded in binary form, using sequences of bits. Understanding this basic premise is crucial, because the way data is structured impacts how it is accessed and read. A byte, composed of eight bits, represents a simple unit of digital information. CDs use this configuration to store complex audio files, which are then decoded by devices like CD players. In practical terms, to negotiate digital information, one needs to understand both the storage capacity and the efficiency with which data can be read and processed by electronic devices.