Analyze the standing sound waves in an open-closed tube to show that the possible wavelengths and frequencies are given by Equation 17.18.

Short Answer

Expert verified

The conditions of having one node and one antinode at the end the formulas that one could get are,

Step by step solution

01

The concept of node and antinode 

Node is a point which's amplitude is 0. The point in which amplitude is maximum is known as antinodes.

02

Finding out the frequencies based on the information 

In an open-closed tube, the presence of node and antinode is a must. so, the possible arrangements can reach to 14λ,34λ,54λ

Thus the outcome is L=mλ4

Here m= 1, 3, 5 and the arrangements will be λm=4Lm

In order to make the connection with frequency, the wave has speed v and wavelength λand the frequency could be found as v=λff=vλ

The substitution of the wavelength isrole="math" localid="1649148923810" fm=mv4L

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 Vaia!

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

Deep-sea divers often breathe a mixture of helium and oxygen to avoid getting the “bends” from breathing high-pressure nitrogen. The helium has the side effect of making the divers’ voices sound odd. Although your vocal tract can be roughly described as an open-closed tube, the way you hold your mouth and position your lips greatly affects the standing-wave frequencies of the vocal tract. This is what allows different vowels to sound different. The “ee” sound is made by shaping your vocal tract to have standing- wave frequencies at, normally, 270 Hz and 2300 Hz. What will these frequencies be for a helium-oxygen mixture in which the speed of sound at body temperature is 750 m/s? The speed of sound in air at body temperature is 350 m/s.

A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightly warm, so the speed of sound is 344 m/s. If the string’s linear density is 0.600 g/m and the tension is 150 N, how long is the vibrating section of the violin string?

The two highest-pitch strings on a violin are tuned to 440 Hz (the A string) and 659 Hz (the E string). What is the ratio of the mass of the A string to that of the E string? Violin strings are all the same length and under essentially the same tension.

FIGURE EX17.27 shows the circular wave fronts emitted by two

wave sources.

a. Are these sources in phase or out of phase? Explain.

b. Make a table with rows labeled P, Q, and R and columns

labeled r1,r2,r,and C/D. Fill in the table for points P, Q, and

R, giving the distances as multiples of l and indicating, with a

C or a D, whether the interference at that point is constructive

or destructive.

Biologists think that some spiders “tune” strands of their web to give an enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20 µm, and spider silk has a density of 1300 kg/m3. To have a fundamental frequency at 100 Hz, to what tension must a spider adjust a 12-cm-long strand of silk?

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.

Sign-up for free