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Chapter 16: Problem 48

A piston is fitted in a cylindrical tube of small cross section with the other end of the tube open. The tube resonates with a tuning fork of frequency \(512 \mathrm{~Hz}\). The piston is gradually pulled out of the tube and it is found that a second resonance occurs when the piston is pulled out through a distance of \(32 \cdot 0 \mathrm{~cm} .\) Calculate the speed of sound in the air of the tube.

Short Answer

Expert verified
The speed of sound in the air is approximately \(327.68 \, \text{m/s}\).

Step by step solution

01

Understand the Problem

We have a cylindrical tube with a piston at one end and open at the other. A tuning fork of frequency \(512 \text{ Hz}\) causes the tube to resonate when the piston is pulled through a certain distance. The goal is to calculate the speed of sound in air.
02

Recall the Resonance Condition

For a tube with one end open and the other end closed (such as our tube with a piston), resonance occurs at odd multiples of the quarter wavelengths. The first resonance occurs at the \(\frac{1}{4}\) wavelength, the second at \(\frac{3}{4}\) wavelength, etc.
03

Calculate the Effective Wavelength

At the first resonance, the length of the air column \(L_1\) represents \(\frac{1}{4}\) of the wavelength \(\lambda\), and at the second resonance, \(L_2 = 32 \, \text{cm}\) represents \(\frac{3}{4}\) of the wavelength. Let \(L_0\) be the end correction for the open end. Thus, \(L_2 = \frac{3\lambda}{4} + L_0\).
04

Solve for the Wavelength

We know the distance the piston was pulled out is \(32.0 \, \text{cm}\). Since this is the difference between \(\frac{3\lambda}{4}\) and \(\frac{1\lambda}{4}\), we can use:\[\frac{3\lambda}{4} - \frac{1\lambda}{4} = 32 \, \text{cm}\] Therefore, \(\frac{2\lambda}{4} = 32 \, \text{cm}\) or \(\frac{\lambda}{2} = 32 \, \text{cm}\), leading to \(\lambda = 64 \, \text{cm} = 0.64 \, \text{m}\).
05

Calculate the Speed of Sound

Using the formula for the speed of sound \(v\), which is \(v = f\lambda\), where \(f\) is the frequency and \(\lambda\) is the wavelength. We have \(f = 512 \, \text{Hz}\) and \(\lambda = 0.64 \, \text{m}\). Substituting the values, we get:\[v = 512 \times 0.64 = 327.68 \, \text{m/s}\].

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Key Concepts

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

Speed of Sound
The speed of sound is a fundamental concept in physics, representing how quickly sound travels through a medium, such as air. When dealing with experiments like resonance in tubes, understanding this speed allows us to calculate other crucial variables. In air, the speed of sound typically ranges around 343 meters per second at room temperature, though various factors such as temperature, humidity, and air pressure can influence this value.
\(v = f\lambda\) is the primary formula used to determine the speed of sound, where \(v\) denotes velocity, \(f\) is the frequency, and \(\lambda\) signifies the wavelength.
  • Factors influencing sound speed include:
    • Temperature: Higher temperatures lead to faster sound speeds.
    • Medium: Sound travels slower in gases than in liquids or solids.
When analyzing resonance in tubes, calculating the speed becomes paramount for accurately assessing subsequent experimental results.
Tuning Fork Frequency
A tuning fork is a tool that vibrates at a specific frequency when struck, and it's widely used in experiments to provide a consistent sound wave. The frequency of a tuning fork is its most essential feature. It determines the pitch of the sound and influences how a tube will resonate with that frequency.
In our scenario, the tuning fork had a frequency of 512 Hz, which serves as the stable frequency source in our resonance tube experiment.
  • Why is frequency important?
    • Defines how many complete vibrations occur per second.
    • Helps determine the wavelength in resonance scenarios.
Understanding the frequency allows students to predict which resonant frequencies are possible within a given tube, depending on the tube's dimensions and conditions.
Wavelength Calculation
Wavelength is the distance between successive crests of a wave and plays a central role in understanding wave behavior. In resonance experiments, calculating the wavelength is a critical step.
Using the formula \(\lambda = \frac{2L}{n}\), where \(L\) is the effective length of the tube and \(n\) represents the harmonic number, helps determine the wavelength of sound in the tube.
  • In our exercise:
    • First resonant length correlates to \(\frac{\lambda}{4}\).
    • Second resonant length correlates to \(\frac{3\lambda}{4}\).
    • Thus, wavelength \(\lambda = 0.64\) m was calculated.
Accurate wavelength calculation enables precision in determining the speed of sound and resonance conditions within the tube.
Quarter Wavelength Resonance
Quarter wavelength resonance refers to the phenomenon where a tube resonates at specific lengths corresponding to odd multiples of a quarter wavelength. In a tube with one end closed and the other open, resonance occurs at these lengths due to the presence of a node at the closed end and an antinode at the open end.
For instance:
  • First resonance occurs at \(\frac{\lambda}{4}\).
  • Second resonance occurs at \(\frac{3\lambda}{4}\).
This pattern repeats as you increase the number of quarter wavelengths within the tube. Understanding this concept:
  • Explains how the effective length of the air column changes with resonance.
  • Assists in calculating wavelength and predicting further resonance points.
By leveraging quarter wavelength resonance, we can solve practical problems related to sound waves and acoustic applications.

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Most popular questions from this chapter

A car moving at \(108 \mathrm{~km} \mathrm{~h}^{-1}\) finds another car in front of it going in the same direction at \(72 \mathrm{~km} \mathrm{~h}^{-1}\). The first car sounds a horn that has a dominant frequency of \(800 \mathrm{~Hz}\). What will be the apparent frequency heard by the driver in the front car ? Speed of sound in air \(=330 \mathrm{~m} \mathrm{~s}^{-1}\).

Two point sources of sound are kept at a separation of \(10 \mathrm{~cm} .\) They vibrate in phase to produce waves of wavelength \(5 \cdot 0 \mathrm{~cm} .\) What would be the phase difference between the two waves arriving at a point \(20 \mathrm{~cm}\) from one source (a) on the line joining the sources and (b) on the perpendicular bisector of the line joining the sources?

A source of sound emitting a \(1200 \mathrm{~Hz}\) note travels along a straight line at a speed of \(170 \mathrm{~m} \mathrm{~s}^{-1}\). A detector is placed at a distance of \(200 \mathrm{~m}\) from the line of motion of the source. (a) Find the frequency of sound received by the detector at the instant when the source gets closest to it. (b) Find the distance between the source and the detector at the instant it detects the frequency \(1200 \mathrm{~Hz}\). Velocity of sound in air \(=340 \mathrm{~m} \mathrm{~s}^{-1}\).

A source of sound operates at \(2 \cdot 0 \mathrm{kHz}, 20 \mathrm{~W}\) emitting sound uniformly in all directions. The speed of sound in air is \(340 \mathrm{~m} \mathrm{~s}^{-1}\) and the density of air is \(1 \cdot 2 \mathrm{~kg} \mathrm{~m}^{-3}\). (a) What is the intensity at a distance of \(6 \cdot 0 \mathrm{~m}\) from the source? (b) What will be the pressure amplitude at this point? (c) What will be the displacement amplitude at this point?

A Kundt's tube apparatus has a copper rod of length \(1 \cdot 0 \mathrm{~m}\) clamped at \(25 \mathrm{~cm}\) from one of the ends. The tube contains air in which the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\). The powder collects in heaps separated by a distance of \(5 \cdot 0 \mathrm{~cm}\). Find the speed of sound waves in copper.
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