/*! 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 19 A thin, \(75.0 \mathrm{~cm}\) wi... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 19

A thin, \(75.0 \mathrm{~cm}\) wire has a mass of \(16.5 \mathrm{~g}\). One end is tied to a nail, and the other end is attached to a screw that can be adjusted to vary the tension in the wire. (a) To what tension (in newtons) must you adjust the screw so that a transverse wave of wavelength \(3.33 \mathrm{~cm}\) makes 625 vibrations per second? (b) How fast would this wave travel?

Short Answer

Expert verified
The tension to adjust the screw to is approximately 94.7 N. The speed of this wave is approximately 20.79 m/s.

Step by step solution

01

Compute the speed of the wave

The relationship between speed (v), frequency (f), and wavelength (位) of a wave is expressed as \(v = f位\). Hence, we first calculate the speed: \(v = 625 s^{-1} * 3.33 * 10^{-2} m = 20.79 ms^{-1}\)
02

Compute the mass per unit length

The mass per unit length (碌) of the wire is computed as the total mass divided by the total length: \(碌 = 16.5g/75.0cm = 2.2 * 10^{-4} kgm^{-1}\)
03

Determine the tension

Knowing the speed (v) and the mass per unit length (碌), we can apply the formula for speed on a stretched string: \(v = \sqrt{T/碌}\), where T is the tension. Solving for T: \(T = 碌v^2 = 2.2 * 10^{-4} kgm^{-1} * (20.79 ms^{-1})^2 鈮� 94.7 N\)
04

Compute the speed of the wave, given tension, mass, and length

With tension (T) and mass per unit length (碌) known, we can determine the speed of the wave using the formula: \(v = \sqrt{T/碌} = \sqrt{94.66N / 2.2 * 10^{-4} kgm^{-1}} 鈮� 20.79 ms^{-1}\).

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Key Concepts

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

Wave speed
Wave speed is an essential concept in understanding how waves travel through a medium. It refers to how fast a wave propagates over time. In this exercise, wave speed was determined using the formula related to frequency and wavelength: \[ v = f \lambda \] where:
  • \( v \) is the wave speed
  • \( f \) is the frequency of the wave, which was 625 vibrations per second in our scenario
  • \( \lambda \) is the wavelength, given as 3.33 cm or 0.0333 meters
Substituting the given values, the wave speed calculated was approximately 20.79 meters per second. This helps us understand how quickly the energy from the wave travels through the string, which is crucial for predicting wave behavior in different physical situations.
Tension in a string
Tension in a string plays a significant role in defining how a wave travels along it. Tension refers to the force applied along the length of the string, stretching it tightly. When a transverse wave travels, this tension determines the wave's speed and stability. By the relation:\[ v = \sqrt{\frac{T}{\mu}} \]- \( T \) represents the tension in the string- \( \mu \) is the mass per unit lengthTo find the tension \( T \), you start from knowing the wave speed and the mass per unit length and rearrange the formula to:\[ T = \mu v^2 \]For the given exercise, substituting the computed mass per unit length \( \mu \) and the calculated wave speed \( v \), the tension amounted to approximately 94.7 Newtons. This shows how crucial adjusting the tension is, for achieving the desired wave properties.
Mass per unit length
The mass per unit length of a string is a fundamental property that affects wave motion. Often symbolized as \( \mu \), it's calculated as the mass divided by the length of the string:\[ \mu = \frac{m}{L} \]In our exercise, the wire had a mass of 16.5 grams and a length of 75.0 cm. Converting these to base units:
  • Mass \( m = 16.5 \) grams \( = 0.0165 \) kg
  • Length \( L = 75.0 \) cm \( = 0.75 \) m
The mass per unit length \( \mu \) was calculated as 0.00022 kg/m. This property directly influences how the wave travels along the wire. The mass per unit length gives us insight into the distribution of mass along the string, which affects the speed of waves moving through it. A smaller \( \mu \) allows waves to propagate faster under the same tension.

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Most popular questions from this chapter

A string or rope will break apart if it is placed under too much tensile stress [see Eq. (11.8)]. Thicker ropes can withstand more tension without breaking because the thicker the rope, the greater the cross-sectional area and the smaller the stress. One type of steel has density \(7800 \mathrm{~kg} / \mathrm{m}^{3}\) and will break if the tensile stress exceeds \(7.0 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\). You want to make a guitar string from \(4.0 \mathrm{~g}\) of this type of steel. In use, the guitar string must be able to withstand a tension of \(900 \mathrm{~N}\) without breaking. Your job is to determine (a) the maximum length and minimum radius the string can have; (b) the highest possible fundamental frequency of standing waves on this string, if the entire length of the string is free to vibrate.

Scale length is the length of the part of a guitar string that is free to vibrate. A standard value of scale length for an acoustic guitar is 25.5 in. The frequency of the fundamental standing wave on a string is determined by the string's scale length, tension, and linear mass density. The standard frequencies \(f\) to which the strings of a sixstring guitar are tuned are given in the table: $$ \begin{array}{l|llllll} \text { String } & \text { E2 } & \text { A2 } & \text { D3 } & \text { G3 } & \text { B3 } & \text { E4 } \\ \hline f(\mathbf{H z}) & 82.4 & 110.0 & 146.8 & 196.0 & 246.9 & 329.6 \end{array} $$ Assume that a typical value of the tension of a guitar string is \(78.0 \mathrm{~N}\) (although tension varies somewhat for different strings). (a) Calculate the linear mass density \(\mu\) (in \(\mathrm{g} / \mathrm{cm}\) ) for the \(\mathrm{E} 2, \mathrm{G} 3\), and \(\mathrm{E} 4\) strings. (b) Just before your band is going to perform, your G3 string breaks. The only replacement string you have is an E2. If your strings have the linear mass densities calculated in part (a), what must be the tension in the replacement string to bring its fundamental frequency to the G3 value of \(196.0 \mathrm{~Hz}\) ?

A \(1.50-\mathrm{m}\) -long rope is stretched between two supports with a tension that makes the speed of transverse waves \(62.0 \mathrm{~m} / \mathrm{s}\). What are the wavelength and frequency of (a) the fundamental; (b) the second overtone; (c) the fourth harmonic?

A \(1.005 \mathrm{~m}\) chain consists of small spherical beads, each with a mass of \(1.00 \mathrm{~g}\) and a diameter of \(5.00 \mathrm{~mm},\) threaded on an elastic strand with negligible mass such that adjacent beads are separated by a center-to-center distance of \(10.0 \mathrm{~mm}\). There are beads at each end of the chain. The strand has a spring constant of \(28.8 \mathrm{~N} / \mathrm{m}\). The chain is stretched horizontally on a frictionless tabletop to a length of \(1.50 \mathrm{~m}\), and the beads at both ends are fixed in place. (a) What is the linear mass density of the chain? (b) What is the tension in the chain? (c) With what speed would a pulse travel down the chain? (d) The chain is set vibrating and exhibits a standing-wave pattern with four antinodes. What is the frequency of this motion? (e) If the beads are numbered sequentially from 1 to \(101,\) what are the numbers of the five beads that remain motionless? (f) The 13th bead has a maximum speed of \(7.54 \mathrm{~m} / \mathrm{s}\). What is the amplitude of that bead's motion? (g) If \(x_{0}=0\) corresponds to the center of the 1 st bead and \(x_{101}=1.50 \mathrm{~m}\) corresponds to the center of the 101 st bead, what is the position \(x_{n}\) of the \(n\) th bead? (h) What is the maximum speed of the 30 th bead?

A small bead of mass \(4.00 \mathrm{~g}\) is attached to a horizontal string. Transverse waves of amplitude \(A=0.800 \mathrm{~cm}\) and frequency \(f=20.0 \mathrm{~Hz}\) are set up on the string. Assume the mass of the bead is small enough that the bead doesn't alter the wave motion. During the wave motion, what is the maximum vertical force that the string exerts on the bead?
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