/*! 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 Violin Strings The heaviest and ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 17: Problem 10

Violin Strings The heaviest and lightest strings on a certain violin have linear densities of \(3.0\) and \(0.29 \mathrm{~g} / \mathrm{m} .\) (a) What is the ratio of the diameter of the heaviest string to that of the lightest string, assuming that the strings are of the same material? (b) What is the ratio of speeds if the strings have the same tension?

Short Answer

Expert verified
The ratio of diameters is approximately 3.22, and the ratio of speeds is approximately 0.31.

Step by step solution

01

Understand the Problem

The problem involves finding the ratio of diameters and speeds of two violin strings, given their linear densities. The strings are made of the same material and assumed to have the same tension.
02

Define Linear Density

Linear density (mass per unit length) is given. For the heaviest string, \(\mu_h = 3.0 \ g/m\), and for the lightest string, \(\mu_l = 0.29 \ g/m\).
03

Relationship Between Linear Density and Diameter

Linear density \(\mu\) can be related to the cross-sectional area and diameter of the string, assuming the material density and shape are constant. \(\mu = \rho A\), where \(A = \frac{\pi d^2}{4}\).
04

Ratio of Diameters

Since the strings are made of the same material, their volume densities \(\rho\) are the same. Therefore, \(\mu_h/\mu_l = d_h^2/d_l^2\). Solving for the diameter ratio: \(\sqrt{\mu_h/\mu_l} = d_h / d_l\). Substituting the given values: \(\sqrt{3.0 / 0.29} = d_h / d_l\). Therefore, \(d_h / d_l = \sqrt{10.34}\).
05

Calculate Diameter Ratio

Compute the ratio: \(d_h / d_l \approx 3.22\).
06

Ratio of Speeds with Same Tension

The speed of a wave on a string is given by \(v = \sqrt{T/\mu}\), where \(T\) is the tension. For the same tension, the speed ratio is \(v_h / v_l = \sqrt{\mu_l / \mu_h}\). Using the given linear densities: \(v_h / v_l = \sqrt{0.29 / 3.0}\).
07

Calculate Speed Ratio

Compute the speed ratio: \(v_h / v_l = \sqrt{0.0967} \approx 0.31\).

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.

linear density
Linear density is a crucial concept, especially in problems involving strings and waves. It refers to the mass per unit length of an object, typically denoted by the symbol \(\mu\). In the context of violin strings, this value is given directly. For example, in our exercise, the heaviest string has a linear density \(\mu_h = 3.0 \, \text{g/m}\), and the lightest string has \(\mu_l = 0.29 \, \text{g/m}\). The linear density can tell us a lot about the string's physical properties and how it will behave when a wave travels through it.
wave speed
Wave speed on a string depends on two main factors: the tension in the string and its linear density. The formula for wave speed \(v\) on a string is \[ v = \sqrt{\frac{T}{\mu}} \], where \(T\) is the tension in the string. When tension is constant across strings, the speed of a wave varies inversely with the square root of the linear density. So, a lighter string (lower \(\mu\)) will generally have a higher wave speed. This is evident in our problem where, given the same tension, the speed of the wave on the lighter string is higher.
material properties
In our problem, it's assumed that both the heaviest and the lightest violin strings are made of the same material. This is important because it means they have the same density \(\rho\). Material properties, like density, play a significant role in determining linear density and the behavior of waves on the string. Given that both strings are made from the same substance, we can directly relate their linear densities to their cross-sectional areas and thus their diameters.
diameter ratio
The diameter of a string is closely related to its linear density when the material is the same. Linear density is linked to diameter via the equation \[ \mu = \rho A \] where \(A = \frac{\pi d^2}{4} \). Since \(\rho \) is constant, comparing the linear densities gives \(\mu_h / \mu_l = d_h^2 / d_l^2\). Solving this ratio for the diameters, it reveals the relationship between the thickness of the strings. In our case: \[ \sqrt{3.0 / 0.29} = d_h / d_l \]. This results in a diameter ratio of approximately 3.22, which means the heaviest string is about 3.22 times thicker than the lightest one.

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

Moving a Nonsymmetric Triangular Pulse A long taut spring is started at a time \(t=0\) with a pulse moving in the \(+x\) direction in the shape given by the function \(f(x)\) with $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{4} x+1 & -4

Resonant Frequencies A string that is stretched between fixed supports separated by \(75.0 \mathrm{~cm}\) has resonant frequencies of 420 and \(315 \mathrm{~Hz}\), with no intermediate resonant frequencies. What are (a) the lowest resonant frequency and (b) the wave speed?

Spring vs. String In this course, we analyzed the motion of a mass on a spring and the oscillations of a taut string. Discuss these two systems, explaining similarities and differences, and give an equation of motion for each.

Equation of a Transverse The equation of a transverse wave traveling along a very long string is \(y(x, t)=(6.0 \mathrm{~cm}) \sin \\{(0.020 \mathrm{\pi} \mathrm{rad} / \mathrm{cm})\) \(x+(4.0 \pi \mathrm{rad} / \mathrm{s}) t]\) where \(x\) and \(y\) are expressed in centimeters and \(t\) is in seconds Determine (a) the amplitude, (b) the wavelength, (c) the frequency, (d) the speed, (e) the direction of propagation of the wave, and (f) the maximum transverse speed of a particle in the string. (g) What is the transverse displacement at \(x=3.5 \mathrm{~cm}\) when \(t=0.26 \mathrm{~s}\) ?

Linear Density The linear density of a string is \(1.6 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\). A transverse wave on the string is described by the equation $$ y(x, t)=(0.021 \mathrm{~m}) \sin [(2.0 \mathrm{rad} / \mathrm{m}) x+(30 \mathrm{rad} / \mathrm{s}) t] $$ What is (a) the wave speed and (b) the tension in the string?
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Astrophysics

Read Explanation

Solid State Physics

Read Explanation

Work Energy and Power

Read Explanation

Fields in Physics

Read Explanation

Dynamics

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.