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A pipe \(90 \mathrm{~cm}\) long is open at both ends. How long must a second pipe, closed at one end, be if it is to have the same fundamental resonance frequency as the open pipe? The two pipes and their fundamental resonances are shown in. As can be seen in the diagram, $$ L_{o}=2\left(\frac{1}{4} \lambda\right) \quad L_{c}=\frac{1}{4} \lambda $$ from which \(L_{c}=\frac{1}{2} L_{o}=45 \mathrm{~cm}\)

Short Answer

Expert verified
The length of the second pipe should be 45 cm.

Step by step solution

01

Understanding the Problem

We have two pipes. One is open at both ends and is 90 cm long, while the other is closed at one end. We need to find the length of the closed pipe such that it has the same fundamental resonance frequency as the open pipe. The length of the open pipe is given as \( L_o = 90 \text{ cm} \), and the length of the closed pipe is given as \( L_c \), which we need to determine.
02

Formula for Open Pipe Resonance

For an open pipe, the fundamental frequency is set by a standing wave where the length of the pipe \( L_o \) is equal to half of the wavelength: \( L_o = \frac{1}{2} \lambda \). Thus, \( \lambda = 2L_o \).
03

Formula for Closed Pipe Resonance

For a closed pipe, the fundamental frequency occurs when the length of the pipe \( L_c \) is a quarter of the wavelength: \( L_c = \frac{1}{4} \lambda \). Thus, \( \lambda = 4L_c \).
04

Matching Wavelengths for Equal Frequency

Since the fundamental resonance frequency must be the same for both pipes, the wavelength of the standing waves must also be the same. Therefore, set the two expressions for \( \lambda \) equal to each other: \( 2L_o = 4L_c \).
05

Solve for the Closed Pipe Length

From the equation \( 2L_o = 4L_c \), solve for \( L_c \) by dividing both sides by 4: \( L_c = \frac{2L_o}{4} = \frac{1}{2}L_o \). Substitute the given value for \( L_o \): \( L_c = \frac{1}{2} \cdot 90 \text{ cm} = 45 \text{ cm} \).

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Key Concepts

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

Open Pipe Resonance
Open pipe resonance occurs when a pipe that is open at both ends supports a standing wave pattern with nodes at both open ends. In an open pipe, the standing wave must fit such that each end of the pipe is a node (point of no displacement). The fundamental frequency of an open pipe resonates when the length of the pipe is equal to half the wavelength of the standing wave. Therefore, we express this relationship as \[L_o = \frac{1}{2} \lambda\]where \(L_o\) is the length of the open pipe and \(\lambda\) is the wavelength of the sound wave. This means the pipe supports the simplest standing wave pattern consisting of one node less than an antinode. The sound propagates in both directions back and forth, creating constructive and destructive interference.
Closed Pipe Resonance
Closed pipe resonance occurs in a pipe that is closed at one end and open at the other. The closed end acts as a node, while the open end acts as an antinode, which is a point of maximum displacement. In the simplest or fundamental resonance mode, the length of the pipe is equal to a quarter of the wavelength. The relation is given by \[L_c = \frac{1}{4} \lambda\]where \(L_c\) is the length of the closed pipe. In this setup, the wave is unable to displace air at the closed end, creating a node, and it achieves maximum displacement at the open end, forming an antinode. Because of the different boundary conditions, closed pipe resonances result in the odd-numbered harmonics only.
Standing Waves
Standing waves are a phenomenon that occurs when two waves of the same frequency and amplitude travel in opposite directions and interfere with each other. This interference pattern creates points along the medium that appear stationary, known as nodes, and points of maximum displacement, known as antinodes.
  • Nodes: Points where the amplitude of vibration is zero.
  • Antinodes: Points where the amplitude of vibration is maximum.
For resonance to occur in pipes, the length of the pipe must support a specific number of nodes and antinodes to satisfy the boundary conditions. The standing waves are crucial for resonance as they determine the specific frequencies, or harmonics, at which the pipe can naturally vibrate.
Fundamental Frequency
The fundamental frequency is the lowest frequency at which an object, such as a pipe, resonates. In terms of musical instruments, it is often referred to as the first harmonic or the base pattern of the standing wave.For open pipes, the fundamental frequency is associated with a wave pattern that fits half of a wavelength inside the length of the pipe. For closed pipes, it aligns with a quarter wavelength fitting inside the pipe length. The fundamental frequency can be calculated using the speed of sound in air and the wavelength:\[f = \frac{v}{\lambda}\]where \(f\) is the fundamental frequency, and \(v\) is the speed of sound.Unlike open pipes that support both odd and even harmonics, closed pipes only support odd harmonics due to their unique boundary conditions. This difference in harmonic structure influences the sound and tone produced by different pipes.

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Most popular questions from this chapter

A string fastened at both ends resonates at \(420 \mathrm{~Hz}\) and \(490 \mathrm{~Hz}\) with no resonant frequencies in between. Find its fundamental resonant frequency. In general, \(f_{n}=n f_{1}\). We are told that \(f_{n}=420 \mathrm{~Hz}\) and \(f_{n+1}=490 \mathrm{~Hz}\). Therefore, $$ 420 \mathrm{~Hz}=n f_{1} \text { and } 490 \mathrm{~Hz}=(n+1) f_{1} $$ Subtract the first equation from the second to obtain \(f_{1}=70.0 \mathrm{~Hz}\).

Suppose that represents a 50 -Hz wave on a string. Take distance \(y_{0}\) to be \(3.0 \mathrm{~mm}\), and distance \(A E\) to be \(40 \mathrm{~cm}\). Find the following for the wave: its \((a)\) amplitude, \((b)\) wavelength, and \((c)\) speed. (a) By definition, the amplitude is distance \(y_{0}\) and is \(3.0 \mathrm{~mm}\). (b) The distance between adjacent crests is the wavelength, and so \(\lambda=20 \mathrm{~cm}\) (c) \(v=\lambda f=(0.20 \mathrm{~m})\left(50 \mathrm{~s}^{-1}\right)=10 \mathrm{~m} / \mathrm{s}\)

A string has its tension doubled; all else kept constant, what happens to the speed of transverse waves that can be set up on the string?

A metal rod \(40 \mathrm{~cm}\) long is dropped, end first, onto a wooden floor and rebounds into the air. Compression waves of many frequencies are thereby set up in the bar. If the speed of compression waves in the bar is \(5500 \mathrm{~m} / \mathrm{s}\), to what lowest- frequency compression wave will the bar resonate as it rebounds? Both ends of the bar will be free, and so antinodes will exist there. In the lowest resonance form (i.e., the one with the longest segments), only one node will exist on the bar, at its center, as illustrated in.We will then have $$ L=2\left(\frac{\lambda}{4}\right) \quad \text { or } \quad \lambda=2 L=2(0.40 \mathrm{~m})=0.80 \mathrm{~m} $$ Then, from \(\lambda=v T=\mathrm{v} / \mathrm{f}\), $$ f=\frac{v}{\lambda}=\frac{5500 \mathrm{~m} / \mathrm{s}}{0.80 \mathrm{~m}}=6875 \mathrm{~Hz}=6.9 \mathrm{kHz} $$

A uniform flexible cable is \(20 \mathrm{~m}\) long and has a mass of \(5.0\) kg. It hangs vertically under its own weight and is vibrated (perpendicularly) from its upper end with a frequency of \(7.0 \mathrm{~Hz}\). (a) Find the speed of a transverse wave on the cable at its midpoint. (b) What are the frequency and wavelength at the midpoint? (b) Because wave crests do not pile up along a string or cable, the number passing one point must be the same as that for any other point. Therefore, the frequency, \(7.0 \mathrm{~Hz}\), is the same at all points. To find the wavelength at the midpoint, we must use the speed we found for that point, \(9.9 \mathrm{~m} / \mathrm{s}\). That gives us $$ \lambda=\frac{v}{f}=\frac{9.9 \mathrm{~m} / \mathrm{s}}{7.0 \mathrm{~Hz}}=1.4 \mathrm{~m} $$

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