/*! 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 28 The rate of vibration of a strin... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 28

The rate of vibration of a string under constant tension varies inversely with the length of the string. If a string is 48 inches long and vibrates 256 times per second, what is the length of a string that vibrates 576 times per second?

Short Answer

Expert verified
The new length of the string is 21.33 inches.

Step by step solution

01

- Understand the relationship

The rate of vibration varies inversely with the length of the string. This means that as the length of the string increases, the rate of vibration decreases, and vice versa. Mathematically, this can be represented as: \[ f \times l = k \] where \( f \) is the frequency, \( l \) is the length of the string, and \( k \) is a constant.
02

- Determine the constant

Using the given values for the first string, substitute \( f = 256 \) vibrations per second and \( l = 48 \) inches into the equation \[ 256 \times 48 = k \]Calculate the constant \( k \): \[ k = 256 \times 48 = 12288 \]
03

- Find the new length

With the constant \( k \) determined, use the given frequency of the new string, \( f = 576 \) vibrations per second. Substitute \( f = 576 \) and \( k = 12288 \) into the equation to find the new length \( l \): \[ 576 \times l = 12288 \] Solve for \( l \): \[ l = \frac{12288}{576} = 21.33 \text{ inches} \]

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 91影视!

Key Concepts

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

Rate of Vibration
The rate of vibration refers to how frequently something oscillates or moves back and forth. In the context of a vibrating string, this is typically measured in vibrations per second, also known as frequency.
When considering the rate of vibration, it is essential to understand that different factors can influence it. In our example with the vibrating string, the length of the string plays a significant role.
As a general rule:
  • If the string is longer, it will vibrate more slowly, thus having a lower frequency.
  • If the string is shorter, it will vibrate more quickly, thus having a higher frequency.
Understanding this fundamental concept helps you grasp why changing the length of the string affects how fast it vibrates.
Constant of Variation
In inverse variation problems, you'll frequently encounter something called the 'constant of variation,' denoted as k.
In this specific exercise, the equation is \( f \times l = k \), where f is the frequency and l is the length. This signifies that when you multiply the frequency by the length of the string, you get a constant value.
Let's break down the steps to identify the constant of variation:
  • You start with the given values. Here, f = 256 vibrations per second, and l = 48 inches.
  • Substitute these into the equation to find k: 256 \times 48 = 12288
  • So, our constant of variation k equals 12288. This value remains constant as long as the tension in the string remains unchanged.
Understanding the constant of variation is crucial, as it allows you to find unknown variables when given different sets of frequency and length.
Frequency and Length Relationship
One of the core ideas in this exercise is understanding the relationship between frequency and length. This concept stems from the idea of inverse variation.
Inverse variation implies that when one quantity increases, the other decreases proportionally. In mathematical terms, we use the formula: \( f \times l = k \)
With this relationship in mind:
  • If you increase the length (l), the frequency (f) will decrease to maintain the constant k.
  • If you decrease the length (l), the frequency (f) will increase.
This relationship helps you to immediately see the balance. In our exercise, after finding the constant k, we could determine the new length of the string by rearranging the equation:
  • Given new frequency f = 576 vibrations per second
  • We used 12288 = 576 \times l
  • Solving for l, we found the new length to be 21.33 inches.
This relationship is vital for solving similar problems regarding the inverse variation of quantities.
Algebraic Equations
Algebraic equations are essential tools in solving problems involving variable relationships. In this exercise, understanding how to manipulate the algebraic equation was key to finding the solution.
We started with the basic equation of inverse variation: \( f \times l = k \)
Here's a step-by-step approach for using algebraic equations:
  • Identify the given values and substitute them into the equation to find the constant, k. For example, 256 \times 48 = 12288.
  • Use the constant in conjunction with the new given value to find the unknown variable. For example, using k = 12288 and f = 576 vibrations per second, we can solve for the new length l.
  • Solve the equation. Rearrange the equation to isolate the unknown variable and perform the calculations: \[ l = \frac{12288}{576} = 21.33\text{ inches} \]
Algebraic equations serve as the backbone for solving such problems. Understanding the steps and keeping the variables in mind will help you efficiently find the solutions.

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

At the corner Shell station, the revenue \(R\) varies directly with the number \(g\) of gallons of gasoline sold. If the revenue is 34.08 when the number of gallons sold is \(12,\) find a linear equation that relates revenue \(R\) to the number \(g\) of gallons of gasoline sold. Then find the revenue \(R\) when the number of gallons of gasoline sold is \(10.5 .\)

(a) find the intercepts of the graph of each equation and (b) graph the equation. $$ 6 x-4 y=24 $$

Access Ramp A wooden access ramp is being built to reach a platform that sits 30 inches above the floor. The ramp drops 2 inches for every 25 -inch run. (a) Write a linear equation that relates the height \(y\) of the ramp above the floor to the horizontal distance \(x\) from the platform. (b) Find and interpret the \(x\) -intercept of the graph of your equation. (c) Design requirements stipulate that the maximum run be 30 feet and that the maximum slope be a drop of 1 inch for each 12 inches of run. Will this ramp meet the requirements? Explain. (d) What slopes could be used to obtain the 30 -inch rise and still meet design requirements?

Find an equation for the line with the given properties. Express your answer using either the general form or the slope-intercept form of the equation of a line, whichever you prefer. Perpendicular to the line \(y=8\); containing the point (3,4)

Are based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Determine the domain of the variable \(x\) in the expression: $$ \sqrt{8-\frac{2}{3} x} $$
See all solutions

Recommended explanations on Math Textbooks

Statistics

Read Explanation

Pure Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation
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.