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Chapter 2: Q1E (page 538)

Question: Example 16.1 (Section 16.1) showed that for sound waves in air with frequency 1000 Hz, a displacement amplitude of \(1.2 \times {10^{ - 8}}\;{\rm{m}}\) produces a pressure amplitude of \(3.0 \times {10^{ - 2}}\;{\rm{Pa}}\). ( a) What is the wavelength of these waves? (b) For 1000-Hz waves in air, what displacement amplitude would be needed for the pressure amplitude to be at the pain threshold, which is 30 Pa? (c) For what wavelength and frequency will waves with a displacement amplitude of \(1.2 \times {10^{ - 8}}\;{\rm{m}}\)produce a pressure amplitude of \(1.5 \times {10^{ - 3}}\;{\rm{Pa}}\)?

Short Answer

Expert verified
  1. The wavelength of these waves is \(0.344\;{\rm{m}}\).

Step by step solution

01

Given data

The given data can be listed below as,

  • The frequency of the sound wave is, \(f = 1000\;{\rm{Hz}}\) .
  • The displacement amplitude is, \(A = 1.2 \times {10^{ - 8}}\;{\rm{m}}\) .
  • The pressure amplitude is, \(p = 3.0 \times {10^{ - 2}}\;{\rm{Pa}}\).
02

Concept

Sound waves are nothing but ripples or disturbances in the medium from which the energy passes through. These disturbances will be observed in a pattern. This pattern is a sound wave. The source of the sound could be an object, equipment, etc.

03

(a) determination of the wavelength

The wavelength of these waves can be calculated using the expression,

\(\lambda = \frac{v}{f}\)

Here v is the velocity of the sound wave, which is 344 m/s.

Substitute the values in the above expression, and we get,

\(\begin{array}{c}\lambda = \frac{{344\;{\rm{m/s}}}}{{1000\;{\rm{Hz}}}}\\ = 0.344\;{\rm{m}}\end{array}\)

Thus, the wavelength of these waves is \(0.344\;{\rm{m}}\) .

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Most popular questions from this chapter

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