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Chapter 16: Problem 6

Sound waves from a loudspeaker spread nearly uniformly in all directions if the wavelength of the sound is much larger than the diameter of the loudspeaker. (a) Calculate the frequency for which the wavelength of sound in air is ten times the diameter of the speaker if the diameter is \(20 \mathrm{~cm} .\) (b) Sound is essentially transmitted in the forward direction if the wavelength is much shorter than the diameter of the speaker. Calculate the frequency at which the wavelength of the sound is one tenth of the diameter of the speaker described above. Take the speed of sound to be \(340 \mathrm{~m} \mathrm{~s}^{-1}\).

Short Answer

Expert verified
a) 170 Hz; b) 17 kHz.

Step by step solution

01

Understanding the Relationship

To find the frequency, we need to remember that the speed of sound in air, denoted as \( v \), is related to the frequency \( f \) and wavelength \( \lambda \) by the equation \( v = f \lambda \). We will use this to solve both parts of the problem.
02

Calculate Wavelength for Uniform Spread

For part (a), the wavelength \( \lambda \) is given to be ten times the diameter of the loudspeaker. With the diameter \( d = 20 \text{ cm} = 0.2 \text{ m} \), the wavelength is \( \lambda = 10 \times 0.2 \text{ m} = 2 \text{ m} \).
03

Determine Frequency for Uniform Spread

Using the equation \( v = f \lambda \), where \( v = 340 \text{ m/s} \) and \( \lambda = 2 \text{ m} \), we can solve for \( f \):\[f = \frac{v}{\lambda} = \frac{340 \text{ m/s}}{2 \text{ m}} = 170 \text{ Hz}\]
04

Calculate Wavelength for Forward Transmission

For part (b), the wavelength \( \lambda \) is one tenth of the diameter. Therefore, \( \lambda = \frac{1}{10} \times 0.2 \text{ m} = 0.02 \text{ m} \).
05

Determine Frequency for Forward Transmission

Using the same relationship \( v = f \lambda \), with \( \lambda = 0.02 \text{ m} \), we find:\[f = \frac{v}{\lambda} = \frac{340 \text{ m/s}}{0.02 \text{ m}} = 17000 \text{ Hz} = 17 \text{ kHz}\]

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Key Concepts

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

Wavelength and Diameter Relationship
When exploring sound waves, the relationship between the wavelength and the diameter of the loudspeaker plays a crucial role in determining how sound spreads. The wavelength is the distance between successive peaks of a wave.
Meanwhile, the diameter of the speaker is the measurement across its broadest part. Understanding how these two dimensions relate helps us comprehend how sound disperses.

  • If the wavelength is much larger than the speaker's diameter, the sound waves can spread almost uniformly in all directions.
  • On the other hand, if the wavelength is shorter than the diameter, the sound tends to be directed more forward.
Recognizing this relationship is essential for accurately predicting whether sound disperses broadly or focuses mainly in one direction.
Speed of Sound
The speed of sound in air is a constant value at given conditions and it profoundly affects how we calculate frequencies and wavelengths. For this problem, we are using a speed of sound value of 340 meters per second (m/s).
This is vital for solving for unknowns in wave properties since the speed of sound links frequency and wavelength through the formula:\[ v = f \lambda \]where:\( v \) is the speed of sound,\( f \) is the frequency,and \( \lambda \) is the wavelength.
  • This formula allows us to determine frequency if the speed and wavelength are known.
  • Similarly, knowing the speed and frequency, we can find the wavelength.
The speed of sound can vary slightly under different atmospheric conditions, such as temperature and pressure, but 340 m/s is a commonly accepted value under standard conditions.
Uniform Sound Wave Spread
When a loudspeaker emits sound, the distribution of sound waves depends on the relationship between wavelength and diameter as mentioned earlier. When the wavelength is much larger, specifically when it is ten times larger than the diameter of the loudspeaker, the sound waves tend to spread uniformly.
This phenomenon means that the sound can be heard evenly in all directions around the loudspeaker.

  • In calculations, if you know that the wavelength should be ten times the diameter, you use this ratio to determine the wavelength based on the speaker size.
  • This influence of the wavelength-to-diameter ratio is what allows for designing sound systems that need to project sound evenly across a space.
Hence, ensuring a wavelength much larger than the loudspeaker diameter aids in achieving a uniform sound wave spread.
Sound Wave Forward Transmission
Sound wave forward transmission occurs when the wavelength is considerably shorter than the diameter of the loudspeaker. In such cases, sound waves tend to travel primarily in the forward direction, focusing the sound output towards a specific area.
This is particularly useful in settings where sound needs to be directed narrowly, such as concerts or public address systems.
  • To achieve this, the wavelength should be significantly shorter than the diameter.
  • Typically, a wavelength that is about one-tenth of the diameter results in this forward-directed sound.
Understanding this concept allows for strategic design in sound systems, creating environments where sound directionality is a priority. This principle is essential for engineers and sound designers when requiring focused sound distribution.

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Most popular questions from this chapter

In Quincke's experiment, the sound intensity has a minimum value \(I\) at a particular position. As the sliding tube is pulled out by a distance of \(16: 5 \mathrm{~mm}\), the intensity increases to a maximum of \(9 I .\) Take the speed of sound in air to be \(330 \mathrm{~m} \mathrm{~s}^{-1}\). (a) Find the frequency of the sound source. (b) Find the ratio of the amplitudes of the two waves arriving at the detector assuming that it does not change much between the positions of minimum intensity and maximum intensity.

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.

A violin player riding on a slow train plays a \(440 \mathrm{~Hz}\) note. Another violin player standing near the track plays the same note. When the two are close by and the train approaches the person on the ground, he hears \(4 \cdot 0\) beats per second. The speed of sound in air= \(340 \mathrm{~m} / \mathrm{s}\). (a) Calculate the speed of the train. (b) What beat frequency is heard by the player in the train?

A source of sound with adjustable frequency produces 2 beats per second with a tuning fork when its frequency is either \(476 \mathrm{~Hz}\) or \(480 \mathrm{~Hz}\). What is the frequency of the tuning fork?

Calculate the speed of sound in oxygen from the following data. The mass of \(22 \cdot 4\) litre of oxygen at STP \(\left(T=273 \mathrm{~K}\right.\) and \(p=1 \cdot 0 \times 10^{5} \mathrm{Nm}^{-2}\) ) is \(32 \mathrm{~g}\), the molar heat capacity of oxygen at constant volume is \(C_{V}=2 \cdot 5\) \(R\) and that at constant pressure is \(C_{p}=3 \cdot 5 R\)
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