/*! 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 30 The water level in a vertical gl... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The water level in a vertical glass tube \(1.00 \mathrm{~m}\) long can be adjusted to any position in the tube. A tuning fork vibrating at \(900 \mathrm{~Hz}\) is held just over the open top end of the tube, to set up a standing wave of sound in the air-filled top portion of the tube. (That airfilled top portion acts as a tube with one end closed and the other end open.) (a) For how many different positions of the water level will sound from the fork set up resonance in the tube's air-filled portion? What are the (b) least and (c) second least water heights in the tube for resonance to occur?

Short Answer

Expert verified
The number of resonant positions is 5; the least water height is 0.14275 m, and the second least is 0.33325 m.

Step by step solution

01

Understanding the Problem

We have a vertical tube filled with water and a tuning fork above it creating sound waves. The tube acts like having one end open and another closed when sound waves are introduced. We need to determine the number of resonant positions, and the least and the second least water height positions for resonance.
02

Resonance in a Tube with One Open End

A tube with one open end and one closed end produces resonances at odd harmonics (1st, 3rd, 5th, ...). The positions are given by \(L = \frac{(2n-1)\lambda}{4}\), where \(n\) is the harmonic number and \(L\) the length of the air column.
03

Calculating Wavelength

The speed of sound in air is approximately 343 m/s. Since the frequency \(f = 900 \mathrm{~Hz}\), we use \(v = f \times \lambda\) to find the wavelength as \(\lambda = \frac{343}{900} \approx 0.381 \text{ m}\).
04

Finding Possible Resonant Positions

Using the formula for resonance \(L_n = \frac{(2n-1)\lambda}{4}\), calculate potential positions for different odd n values while ensuring that \(L_n \leq 1 \) meter. Compute until the length exceeds the maximum possible 1 meter air column.
05

Compute the Wavelength Positions

For \(n = 1\), \(L_1 = \frac{\lambda}{4} = \frac{0.381}{4} = 0.09525 \text{ m}\); for \(n = 3\), \(L_3 = \frac{3\lambda}{4} = 0.28575 \text{ m}\),for \(n = 5\), \(L_5 = \frac{5\lambda}{4} = 0.47625 \text{ m}\);for \(n = 7\), \(L_7 = \frac{7\lambda}{4} = 0.66675 \text{ m}\), so far still less than 1 m; for \(n = 9\), \(L_9 = \frac{9\lambda}{4} = 0.85725 \text{ m}\). Try \(n = 11\) but it exceeds.
06

Convert Positions to Water Levels

Since the water level must account for the entire tube minus the air length, calculate water level heights:For \(L_1\), height \(1 - 0.09525 = 0.90475\text{ m}\);For \(L_3\), height \(1 - 0.28575 = 0.71425\text{ m}\);For \(L_5\), height \(1 - 0.47625 = 0.52375\text{ m}\).For \(L_7\), height \(1 - 0.66675 = 0.33325\text{ m}\). For \(L_9\), height \(1 - 0.85725 = 0.14275\text{ m}\).
07

Final Count of Total Positions

Determine all harmonic numbers resulting in a resonant length \(L\) less than or equal to 1 meter.

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.

Resonance
Resonance occurs when an object vibrates at a frequency that matches the natural frequency of another object, causing it to vibrate as well. In the context of a tube with a tuning fork, resonance happens when the frequency of the sound wave matches one of the natural frequencies of the air column inside the tube. This creates standing waves where certain points called nodes (with no movement) and antinodes (with maximum movement) are formed.
For a tube with one open end and one closed end, like in the exercise, resonance is set up at odd harmonics (1st, 3rd, 5th, etc.). These standing wave patterns allow sound waves to reinforce each other, amplifying the sound at these specific lengths.
Resonance is an essential concept because it explains how sound can be amplified in musical instruments and in engineering applications. When an instrument or a structure resonates, it can produce loud and rich sounds or beneficial vibrational properties.
Harmonics
Harmonics are integral multiples of the fundamental frequency of a system. They determine the pattern and positions where standing waves occur in the air column. In a tube with one end closed, which resembles the setup in the exercise, only odd harmonics are allowed.
  • 1st Harmonic (=1): The simplest standing wave, where the air column length is \(\frac{\lambda}{4}\).
  • 3rd Harmonic (=3): Consists of 3/4th the wavelength in the column.
  • 5th Harmonic (=5): Involves 5/4th of the wavelength.
  • And so on.

These harmonics directly dictate where resonance, and consequently constructive interference, can occur, amplifying sound at each harmonic frequency. The natural resonant frequencies are crucial for designing musical instruments, as they determine the notes that can be played.
Speed of Sound
The speed of sound is a fundamental property, determining how fast sound waves travel through a medium, such as air. At room temperature, this speed is approximately 343 m/s in air. In the problem, this value facilitates the calculation of the sound wavelength when the frequency of the tuning fork, 900 Hz, is known.
The formula to relate speed, wavelength, and frequency is given by \( v = f \cdot \lambda \) where \( v \) is the speed of sound, \( f \) is the frequency, and \( \lambda \) is the wavelength. Here, knowing the speed and frequency, we derive the wavelength as \( \lambda = \frac{v}{f} = \frac{343 \text{ m/s}}{900 \text{ Hz}} \approx 0.381 \text{ m} \).
The speed of sound can vary based on factors such as temperature, humidity, and air pressure, but the given speed provides a standard reference point for the calculations in most exercises.
Wavelength
The wavelength of a sound wave is the physical distance between two successive peaks (or troughs) of the wave. It is crucial as it determines the length of the air column required for each resonant frequency in the tube. In the given problem, the wavelength was calculated using the speed of sound and the frequency of the source.
When standing waves are formed in a tube, the relationship between the wavelength and the air column length is governed by the harmonic number. For example, the first harmonic occurs at a length equal to \(\frac{\lambda}{4}\), the third at \(\frac{3\lambda}{4}\), and so on. These small fractions of the wavelength are used to find the specific positions inside the tube where resonance occurs.
Understanding wavelengths allows for the designing of various acoustic and audio devices to enhance or reduce sound wave propagation and to align with the desired resonant frequencies.

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

A girl is sitting near the open window of a train that is moving at a velocity of \(10.00 \mathrm{~m} / \mathrm{s}\) to the east. The girl's uncle stands near the tracks and watches the train move away. The locomotive whistle emits sound at frequency \(520.0 \mathrm{~Hz}\). The air is still. (a) What frequency does the uncle hear? (b) What frequency does the girl hear? A wind begins to blow from the east at \(10.00 \mathrm{~m} / \mathrm{s}\). (c) What frequency does the uncle now hear? (d) What frequency does the girl now hear?

When the door of the Chapel of the Mausoleum in Hamilton, Scotland, is slammed shut, the last echo heard by someone standing just inside the door reportedly comes \(15 \mathrm{~s}\) later. (a) If that echo were due to a single reflection off a wall opposite the door, how far from the door is the wall? (b) If, instead, the wall is \(32.0 \mathrm{~m}\) away, how many reflections (back and forth) occur?

An acoustic burglar alarm consists of a source emitting waves of frequency \(30.0 \mathrm{kHz}\). What is the beat frequency between the source waves and the waves reflected from an intruder walking at an average speed of \(0.950 \mathrm{~m} / \mathrm{s}\) directly away from the alarm?

Pipe \(A\), which is \(1.80 \mathrm{~m}\) long and open at both ends, oscillates at its third lowest harmonic frequency. It is filled with air for which the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). Pipe \(B\), which is closed at one end, oscillates at its second lowest harmonic frequency. This frequency of \(B\) happens to match the frequency of \(A\). An \(x\) axis extends along the interior of \(B\), with \(x=0\) at the closed end. (a) How many nodes are along that axis? What are the (b) smallest and (c) second smallest value of \(x\) locating those nodes? (d) What is the fundamental frequency of \(B\) ?

A point source emits sound waves isotropically. The intensity of the waves \(6.00 \mathrm{~m}\) from the source is \(4.50 \times 10^{-4} \mathrm{~W} / \mathrm{m}^{2}\). Assuming that the energy of the waves is conserved, find the power of the source.

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.