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Chapter 10: Problem 1465

Tube \(A\) has both ends open while tube \(B\) has one end closed, otherwise they are identical. The ratio of fundamental frequency of tube \(\mathrm{A}\) and \(\mathrm{B}\) is \(\ldots \ldots \ldots\) (A) \(1: 2\) (B) \(1: 4\) (C) \(2: 1\) (D) \(4: 1\)

Short Answer

Expert verified
The ratio of the fundamental frequency of Tube A and Tube B is 2:1.

Step by step solution

01

Formula for fundamental frequency of a tube

: For a tube with both ends open or both ends closed: \(f_{1}=\dfrac{v}{2L}\) For a tube with one end open, and one end closed: \(f_{2}=\dfrac{v}{4L}\) where \(f_{1}\) is the fundamental frequency of an open-closed tube, \(f_{2}\) is the fundamental frequency of an open-open tube, \(L\) is the length of the tube, and \(v\) is the speed of sound in air.
02

Evaluate the fundamental frequency of Tube A (both ends open)

: Using the formula for tubes with both ends open: \(f_{1}=\dfrac{v}{2L}\)
03

Evaluate the fundamental frequency of Tube B (one end closed)

: Using the formula for tubes with one end open: \(f_{2}=\dfrac{v}{4L}\)
04

Find the ratio of fundamental frequencies

: Now we will find the ratio of the fundamental frequencies of Tube A and Tube B: \(\dfrac{f_{1}}{f_{2}}=\dfrac{\dfrac{v}{2L}}{\dfrac{v}{4L}}\) The 'v' and 'L' get canceled out, leaving: \(\dfrac{f_{1}}{f_{2}}=\dfrac{2}{1}\) Thus, the ratio of fundamental frequency of Tube A to Tube B is 2:1, which corresponds to option (C). Answer: \(\boxed{\text{(C)}\ 2:1}\).

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Key Concepts

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

fundamental frequency
The fundamental frequency refers to the lowest frequency produced by any vibrating object, like a tube, and it determines the pitch of the sound. For tubes, this frequency depends on several factors including the length of the tube and whether the ends are open or closed.
In simple terms, the vibration inside the tube creates sound waves that resonate at this fundamental frequency.
  • For an open-open tube, the sound can easily escape from both ends, so the fundamental frequency is higher.
  • For an open-closed tube, sound is reflected back at the closed end, affecting the frequency produced.
Understanding the fundamental frequency is crucial when predicting how a tube will sound, since it is the base upon which other harmonics or overtones are built.
open-closed tube
An open-closed tube has one open end and one closed end. This creates a specific pattern of resonating sound waves.
When sound waves travel through the tube, they reflect off the closed end, creating a node, and form an antinode at the open end. This pattern alters the sound characteristics significantly.
The fundamental frequency for an open-closed tube can be calculated using the formula:\( f_{2} = \frac{v}{4L} \)where
  • \(v\) is the speed of sound in air.
  • \(L\) is the length of the tube.
This indicates that the wavelength is four times the length of the tube for the first harmonic, leading to a lower fundamental frequency compared to an open-open tube of the same length. Understanding this behavior is critical when designing musical instruments that employ open-closed tubes, such as some wind instruments.
open-open tube
An open-open tube is open at both ends, allowing air and sound to move freely. This type of tube supports different modes of vibration compared to an open-closed tube.
The sound waves create antinodes at both openings, which means the tube resonates with a different harmonic series.
The fundamental frequency for an open-open tube can be calculated using the formula:\( f_{1} = \frac{v}{2L} \)Here,
  • \(v\) is the speed of sound in air.
  • \(L\) represents the length of the tube.
This results in a wavelength that is two times the length of the tube. Due to the placement of nodes and antinodes, the fundamental frequency is higher than in an open-closed tube of the same length. This understanding aids in crafting instruments like flutes or organ pipes, which rely on these principles for producing sound.
speed of sound
The speed of sound is a crucial factor when determining the fundamental frequency of sound waves in tubes. It is the speed at which sound travels through the air and is influenced by conditions such as temperature and air pressure.
Typically, the speed of sound in air at room temperature is about 343 meters per second. This velocity affects the frequency and wavelength of the sound produced in the tube.
When calculating frequencies using the formulas:
  • For an open-open tube: \( f_{1} = \frac{v}{2L} \)
  • For an open-closed tube: \( f_{2} = \frac{v}{4L} \)
\(v\) is used as the speed of sound. Knowing \(v\) helps predict how an instrument will sound and allows for precise tuning when altering conditions such as temperature. Understanding these dynamics is essential in practical applications such as music, engineering, and physics.

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Most popular questions from this chapter

Two wires made up of same material are of equal lengths but their radii are in the ratio \(1: 2\). On stretching each of these two strings by the same tension, the ratio between their fundamental frequency is \(\ldots \ldots \ldots .\) (A) \(1: 2\) (B) \(2: 1\) (C) \(1: 4\) (D) \(4: 1\)

When a block of mass \(\mathrm{m}\) is suspended from the free end of a massless spring having force constant \(\mathrm{k}\), its length increases by y. Now when the block is slightly pulled downwards and released, it starts executing S.H.M with amplitude \(\mathrm{A}\) and angular frequency \(\omega\). The total energy of the system comprising of the block and spring is \(\ldots \ldots \ldots\) (A) \((1 / 2) \mathrm{m} \omega^{2} \mathrm{~A}^{2}\) (B) \((1 / 2) m \omega^{2} A^{2}+(1 / 2) \mathrm{ky}^{2}\) (C) \((1 / 2) \mathrm{ky}^{2}\) (D) \((1 / 2) m \omega^{2} A^{2}-(1 / 2) k y^{2}\)

A tuning fork of frequency \(480 \mathrm{~Hz}\) produces 10 beats/s when sounded with a vibrating sonometer string. What must have been the frequency of the string if a slight increase in tension produces fewer beats per second than before? (A) \(480 \mathrm{~Hz}\) (B) \(490 \mathrm{~Hz}\) (C) \(460 \mathrm{~Hz}\) (D) \(470 \mathrm{~Hz}\)

The equation for displacement of a particle at time \(t\) is given by the equation \(\mathrm{y}=3 \cos 2 \mathrm{t}+4 \sin 2 \mathrm{t}\). The frequency of the particle is \(\ldots \ldots \mathrm{s}^{-1}\). (A) \((1 / \pi)\) (B) \(\pi\) (C) \((1 / 2 \pi)\) (D) \((\pi / 2)\)

What should be the speed of a source of sound moving towards a stationary listener, so that the frequency of sound heard by the listener is double the frequency of sound produced by the source? \\{Speed of sound wave is \(\mathrm{v}\\}\) (A) \(\mathrm{v}\) (B) \(2 \mathrm{v}\) (C) \(\mathrm{v} / 2\) (D) \(\mathrm{v} / 4\)
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