The voice of a person who has inhaled helium sounds very high-pitched. Why?

Short Answer

Expert verified

With helium, the speed of sound is high and hence the frequency. Therefore, the pitch of the wave becomes high.

Step by step solution

01

Write the concept of sound producing mechanism 

The sound-producing mechanism is based on resonating cavities, like the throat. The relatively fixed geometry of these cavities determines the relatively fixed wavelengths of sound. The expression for the frequency can be written as:

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

Here, v is the speed of the wave and\(\lambda \)is wavelength of the sound wave.

02

Comparison of the speed of sound in helium and air

The speed of the sound wave depends on the gas-filled in cavities. The density of helium is much less than that of air; therefore, the sound's speed is three times as large as in the air. Thus, the person's frequency will go up by a factor of three, so the person sounds high-pitched.

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Most popular questions from this chapter

Question: The intensity at the threshold of hearing for the human ear at a frequency of about \({\bf{1000}}\,{\bf{Hz}}\)is \({{\bf{I}}_{\bf{0}}}{\bf{ = 1}}{\bf{.0 \times 1}}{{\bf{0}}^{{\bf{ - 12}}}}\,{\bf{W/}}{{\bf{m}}^{\bf{2}}}\), for which\({\bf{\beta }}\), the sound level, is \({\bf{0}}\,{\bf{dB}}\). The threshold of pain at the same frequency is about\({\bf{120}}\,{\bf{dB}}\), or \({\bf{I = 1}}{\bf{.0}}\,{\bf{W/}}{{\bf{m}}^{\bf{2}}}\)corresponding to an increase of intensity by a factor of \({\bf{1}}{{\bf{0}}^{{\bf{12}}}}\)By what factor does the displacement amplitude,\({\bf{A}}\), vary?

How far from the mouthpiece of the flute in Example 12–11 should the hole be that must be uncovered to play \({\bf{F}}\) above middle \({\bf{C}}\) at \({\bf{349}}\;{\bf{Hz}}\)?

Question: (I) A certain dog whistle operates at 23.5 kHz, while another (brand X) operates at an unknown frequency. If humans can hear neither whistle when played separately, but a shrill whine of frequency 5000 Hz occurs when they are played simultaneously, estimate the operating frequency of brand X.

Question: (III) When a player’s finger presses a guitar string down onto a fret, the length of the vibrating portion of the string is shortened, thereby increasing the string’s fundamental frequency (see Fig. 12–36). The string’s tension and mass per unit length remain unchanged. If the unfingered length of the string is l= 75.0 cm, determine the positions x of the first six frets, if each fret raises the pitch of the fundamental by one musical note compared to the neighboring fret. On the equally tempered chromatic scale, the ratio of frequencies of neighboring notes is 21/12.

Figure 12-36

What is the intensity of a sound at the pain level of 120 dB? Compare it to that of a whisper at 20 dB.

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