/*! 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 12 The mass of a string is \(5.0 \t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The mass of a string is \(5.0 \times 10^{-3} \mathrm{kg},\) and it is stretched so that the tension in it is \(180 \mathrm{N}\). A transverse wave traveling on this string has a frequency of \(260 \mathrm{Hz}\) and a wavelength of \(0.60 \mathrm{m} .\) What is the length of the string?

Short Answer

Expert verified
The length of the string is approximately 0.675 m.

Step by step solution

01

Identify the Wave Speed

The wave speed \( v \) of a wave is given by the formula \( v = f \lambda \), where \( f \) is the frequency of the wave, and \( \lambda \) is the wavelength. Substitute the given frequency \( f = 260 \text{ Hz} \) and wavelength \( \lambda = 0.60 \text{ m} \) into the equation to find the wave speed.\[v = 260 \times 0.60 = 156 \text{ m/s}.\]
02

Use the Wave Speed Formula for Strings

The speed of a wave on a string is also given by \( v = \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension in the string, and \( \mu \) is the linear mass density. Rearrange the formula to find the linear mass density \( \mu \):\[\mu = \frac{T}{v^2}.\]Substitute the values \( T = 180 \text{ N} \) and \( v = 156 \text{ m/s} \) into it:\[\mu = \frac{180}{(156)^2} \approx 0.00740 \text{ kg/m}.\]
03

Determine the Length of the String

The linear mass density \( \mu \) is defined as \( \mu = \frac{m}{L} \), where \( m \) is the mass of the string and \( L \) is its length. Rearrange to solve for \( L \):\[L = \frac{m}{\mu}.\]Substitute the mass \( m = 5.0 \times 10^{-3} \text{ kg} \) and the calculated linear mass density \( \mu = 0.00740 \text{ kg/m} \) into the equation:\[L = \frac{5.0 \times 10^{-3}}{0.00740} \approx 0.675 \text{ m}.\]
04

Conclusion: Result

The length of the string is approximately \(0.675 \text{ m}\).

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.

Tension in String
Tension in a string plays a crucial role in determining how waves travel through it. Imagine a guitar string; the tension is like how tightly the string is pulled. Higher tension results in faster wave speeds. This is because the tightness reduces slack, allowing the wave to move more swiftly.
Tension is denoted by the symbol \( T \) and measured in Newtons (N). In the exercise, the string is under a tension of \( 180 \text{ N} \).
When waves travel through a string, they rely on the tension to maintain consistent speed and energy. This means:
  • Higher tension increases the speed of the wave.
  • The wave can travel longer distances without losing energy.
Understanding the relationship between tension and wave speed is key to analyzing how waves behave in various string instruments or any objects with similar dynamics.
Wavelength
Wavelength represents the distance between two consecutive crests or troughs in a wave. It's like measuring the length of one complete wave cycle. In the context of a string, this tells us how spread out the waves are as they travel.
In the exercise, the wavelength \( \lambda \) is given as \( 0.60 \text{ m} \).
Wavelength influences:
  • The appearance and spacing of waves along the string.
  • The resonant frequencies that the string can naturally support.
If you're playing with the tension or changing the length, you are effectively altering the wavelength. Hence, it's essential to consider wavelength when dealing with wave physics in strings.
Frequency
Frequency in physics refers to how many waves pass through a point in one second. It's an indicator of how "fast" the waves are oscillating along the string.
In this example, the frequency \( f \) is \( 260 \text{ Hz} \), meaning \( 260 \) waves pass a point each second.
Frequency is directly related to:
  • The pitch of sound when the wave is audible. For example, higher frequency means a higher pitch.
  • The temporal resolution of vibration, with higher frequencies depicting quicker transitions.
Knowing the frequency is vital for applications like sound engineering and designing musical instruments. It's a central concept for harmonics where each frequency corresponds to a particular note or tone.
Linear Mass Density
Linear mass density \( \mu \) is the mass of the string per unit length. It essentially tells us how "heavy" the string is per meter.
From our solution, we have a linear mass density \( \mu \approx 0.00740 \text{ kg/m} \). This means for every meter of length, the string has around \( 0.00740 \text{ kg} \) of mass.
The linear mass density impacts:
  • The speed at which waves travel through the string, with a lower density allowing faster speeds.
  • The resonant frequencies, affecting the sound the string produces when vibrated.
If you have two strings, one thicker than the other but both are under similar tension, the thicker string will have a higher linear mass density and will generally produce lower frequencies or notes.

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 what temperature is the speed of sound in helium (ideal gas, \(\gamma=1.67,\) atomic mass \(=4.003\) u) the same as its speed in oxygen at \(0^{\circ} \mathrm{C} ?\)

To navigate, a porpoise emits a sound wave that has a wavelength of \(1.5 \mathrm{cm} .\) The speed at which the wave travels in seawater is \(1522 \mathrm{m} / \mathrm{s} .\) Find the period of the wave.

Two blocks are connected by a wire that has a mass per unit length of \(8.50 \times 10^{-4} \mathrm{kg} / \mathrm{m} .\) One block has a mass of \(19.0 \mathrm{kg},\) and the other has a mass of \(42.0 \mathrm{kg} .\) These blocks are being pulled across a horizontal frictionless floor by a horizontal force \(\overrightarrow{\mathbf{P}}\) that is applied to the less massive block. A transverse wave travels on the wire between the blocks with a speed of \(352 \mathrm{m} / \mathrm{s}\) (relative to the wire). The mass of the wire is negligible compared to the mass of the blocks. Find the magnitude of \(\overrightarrow{\mathbf{P}}\)

A source of sound is located at the center of two concentric spheres, parts of which are shown in the drawing. The source emits sound uniformly in all directions. On the spheres are drawn three small patches that may or may not have equal areas. However, the same sound power passes through each patch. The source produces \(2.3 \mathrm{W}\) of sound power, and the radii of the concentric spheres are \(r_{A}=0.60 \mathrm{m}\) and \(r_{\mathrm{B}}=0.80 \mathrm{m}\) (a) Determine the sound intensity at each of the three patches. (b) Thesound power that passes through each of the patches is \(1.8 \times 10^{-3} \mathrm{W}\). Find the area of each patch.

A woman stands a distance \(d\) from a loud motor that emits sound uniformly in all directions. The sound intensity at her position is an uncomfortable \(3.2 \times 10^{-3} \mathrm{W} / \mathrm{m}^{2} .\) There are no reflections. At a position twice as far from the motor, what are (a) the sound intensity and (b) the sound intensity level relative to the threshold of hearing?

See all solutions

Recommended explanations on Physics 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.