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Chapter 12: Q79GP (page 328)

Question: A tuning fork is set into vibration above a vertical open tube filled with water (Fig 12-40). The water level is allowed to drop slowly. As it does so, the air in the tube above the water level is heard to resonate with the tuning fork when the distance from the tube opening to the water level is 0.125 m and again at 0.395 m. What is the frequency of the tuning fork?

FIGURE 12–40 Problem 79.

Short Answer

Expert verified

The frequency of the tuning fork is \(635\;{\rm{Hz}}\).

Step by step solution

01

Determination of frequency of tuning fork

The distance between the two nodes is half of the wavelength. Using this concept, the wavelength is determined, hence the tuning fork's frequency.

02

Given information

The length of the tube is 0.395 m.

The water level is 0.125 m below the top.

03

Find the wavelength of the tuning fork

The distance between the node is one-half of the length. Therefore,

\(\begin{array}{c}\Delta l = \frac{1}{2}\lambda \\\lambda = 2\Delta l\\ = 2\left( {0.395\,\,{\rm{m}} - 0.125\,\,{\rm{m}}} \right)\\ = 0.540\;{\rm{m}}\end{array}\)

04

Find the frequency of the tuning fork

The value of frequency can be calculated as:

Thus, the frequency of the tuning fork is \(635\;{\rm{Hz}}\).

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Most popular questions from this chapter

Question: A dramatic demonstration, called “singing rods,” involves a long, slender aluminum rod held in the hand near the rod’s midpoint. The rod is stroked with the other hand. With a little practice, the rod can be made to “sing,” or emit a clear, loud, ringing sound. For an \({\bf{80}}\)-cm-long rod, (a) what is the fundamental frequency of the sound? (b) What is its wavelength in the rod, and (c) what is the traveling wavelength of the sound in air at\({\bf{2}}{{\bf{0}}^{\bf{o}}}{\bf{C}}\)?

A particular organ pipe can resonate at \({\bf{264}}\,{\bf{Hz}}\), \({\bf{440}}\,{\bf{Hz}}\), and \({\bf{616}}\,{\bf{Hz}}\), but not at any other frequencies in between. (a) Show why this is an open or a closed pipe. (b) What is the fundamental frequency of this pipe?

To make a given sound seem twice as loud, how should a musician change the intensity of the sound?

(a) Double the intensity.

(b) Halve the intensity.

(c) Quadruple the intensity.

(d) Quarter the intensity.

(e) Increase the intensity by a factor of 10.

A noisy truck approaches you from behind a building. Initially you hear it but cannot see it. When it emerges and you do see it, its sound is suddenly “brighter”—you hear more of the high-frequency noise. Explain. [Hint: See Section 11–14 on diffraction.]

A musical note that is two octaves higher than a second note

(a) has twice the frequency of the second note.

(b) has four times the frequency of the second note.

(c) has twice the amplitude of the second note.

(d) is 3 dB louder than the second note.

(e) None of the above.
See all solutions

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