/*! 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 61 For a string stretched between t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For a string stretched between two supports, two succes- sive standing-wave frequencies are 525 Hz and 630 \(\mathrm{Hz}\) . There are other standing-wave frequencies lower than 525 \(\mathrm{Hz}\) and higher than 630 \(\mathrm{Hz}\) . If the speed of transverse waves on the string is \(384 \mathrm{m} / \mathrm{s},\) what is the length of the string? Assume that the mass of the wire is small enough for its effect on the tension in the wire to be neglected.

Short Answer

Expert verified
The length of the string is approximately 1.83 meters.

Step by step solution

01

Understanding Standing Wave Frequencies

Standing wave frequencies on a string fixed at both ends are given by \(f_n = n \frac{v}{2L}\), where \(n\) is an integer, \(v\) is the speed of the wave, and \(L\) is the length of the string. The difference in successive frequencies is \(\Delta f = \frac{v}{2L}\).
02

Calculating the Difference of Frequencies

The difference between the two successive standing wave frequencies provided is \(630 \, \text{Hz} - 525 \, \text{Hz} = 105 \, \text{Hz}\). This difference is \(\Delta f = 105 \, \text{Hz}\).
03

Setting Up the Equation for the Speed of the Wave

Knowing that \(\Delta f = \frac{v}{2L}\), we use the known values: \(\Delta f = 105 \, \text{Hz}\) and \(v = 384 \, \text{m/s}\). We have the equation \(105 = \frac{384}{2L}\).
04

Solving for the Length of the String

Rearrange the equation \(105 = \frac{384}{2L}\) to solve for \(L\). This gives \(L = \frac{384}{2 \times 105}\).
05

Calculating the Final Value

Compute \(L = \frac{384}{210}\), which simplifies to \(L = 1.8286\). Since measurements are often rounded, \(L \approx 1.83 \, \text{meters}\).

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.

Wave Frequency
Wave frequency refers to the number of waves that pass a point in one second. It is measured in Hertz (Hz). Wave frequencies on a string are particularly interesting because they allow us to observe standing waves. Standing waves occur when the frequency of incident waves matches with the natural frequency of the system. This results in a pattern that seems still, with nodes (points of no movement) and antinodes (points of maximum displacement).
In strings fixed at both ends, such as musical instruments, these standing wave frequencies are observed as specific harmonics. Each harmonic corresponds to a different integer value, denoted as \(n\), which represents the mode number. The fundamental frequency (the first harmonic) is the lowest frequency at which the system naturally oscillates and is crucial in determining the pitch of the sound produced.
When calculating the frequency of standing waves, the equation \(f_n = n \frac{v}{2L}\) is used. Here, \(f_n\) represents the frequency of the nth harmonic, \(v\) stands for the wave velocity, and \(L\) is the length of the string. Understanding these frequencies is key to solving problems related to how different wave characteristics influence each other.
Transverse Waves
Transverse waves are types of waves where the motion of the medium is perpendicular to the direction of the wave. For a string, imagine how moving it up and down creates waves that travel horizontally along the string. This perpendicular motion is what defines transverse waves.
These waves are often seen in vibrations of strings in musical instruments, where the displacement of the strings creates sound through vibration of the surrounding air. Thus, transverse waves are foundational in producing music and understanding acoustics.
In the context of this exercise, the speed of transverse waves on a string is provided and plays a vital role in determining string length and resonant frequencies. The speed \(v\) of such waves can be influenced by factors such as tension in the string and its linear density. However, in our exercise, we assume the mass of the wire is negligible, focusing instead on the calculated speed provided as \(384 \, \text{m/s}\). This speed is essential for calculating wave frequencies and solving related problems.
String Length Calculation
Calculating the length of a string for creating standing wave patterns involves using the relationship between wave speed, frequency, and string length. The given exercise provides the frequencies of two successive harmonics and their difference helps us solve for the string length.
First, the difference between successive harmonic frequencies gives us \(\Delta f = \frac{v}{2L}\). In our problem, we've calculated this difference as \(105 \, \text{Hz}\), so we can use this equation to solve for \(L\). This is crucial because the length of the string impacts the frequencies of the harmonics that can be produced, affecting the sound characteristics.
Rearranging the equation \(105 = \frac{384}{2L}\), we isolate \(L\) to find \(L = \frac{384}{2 \times 105}\). Performing the calculation, we simplify to \(L = 1.8286 \, \text{meters}\), then round to \(L \approx 1.83 \, \text{meters}\). This provides us with the string length, demonstrating how fundamental calculations are necessary to determine the characteristics of a string's wave pattern.

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

One string of a certain musical instrument is 75.0 \(\mathrm{cm}\) long and has a mass of 8.75 g. It is being played in a room where the speed of sound is 344 \(\mathrm{m} / \mathrm{s}\) . (a) To what tension must you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.765 \(\mathrm{m} ?\) (Assume that the breaking stress of the wire is very large and isn't exceeded.) (b) What frequency sound does this string produce in its fundamental mode of vibration?

Tuning an Instrument. A musician tunes the C-string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 \(\mathrm{m}\) long and has a mass of 14.4 \(\mathrm{g} .\) (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 \(\mathrm{Hz}\) to 73.4 \(\mathrm{Hz}\) , corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D} ?\)

CALC Ant Joy Ride. You place your pet ant Klyde (mass \(m )\) on top of a horizontal, stretched rope, where he holds on tightly. The rope has mass \(M\) and length \(L\) and is under tension \(F .\) You start a sinusoidal transverse wave of wavelength \(\lambda\) and amplitude \(A\) propagating along the rope. The motion of the rope is in a vertical plane. Klyde's mass is so small that his presence has no effect on the propagation of the wave. (a) What is Klyde's top speed as he oscillates up and down? (b) Klyde enjoys the ride and begs for more. You decide to double his top speed by changing the tension while keeping the wavelength and amplitude the same. Should the tension be increased or decreased, and by what factor?

Transverse waves on a string have wave speed 8.00 \(\mathrm{m} / \mathrm{s}\) , amplitude \(0.0700 \mathrm{m},\) and wavelength 0.320 \(\mathrm{m} .\) The waves travel in the \(-x\) -direction, and at \(t=0\) the \(x=0\) end of the string has its maximum upward displacement. (a) Find the frequency, period, and wave number of these waves. (b) Write a wave function describing the wave. (c) Find the transverse displacement of a par-ticle at \(x=0.360 \mathrm{m}\) at time \(t=0.150 \mathrm{s}\) . (d) How much time must elapse from the instant in part (c) until the particle at \(x=0.360 \mathrm{m}\) next has maximum upward displacement?

Energy Output. By measurement you determine that sound waves are spreading out equally in all directions from a point source and that the intensity is 0.026 \(\mathrm{W} / \mathrm{m}^{2}\) at a distance of 4.3 \(\mathrm{m}\) from the source. (a) What is the intensity at a distance of 3.1 m from the source? (b) How much sound energy does the source emit in one hour if its power output remains constant?

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.