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Anettt [7]
3 years ago
10

The harmonic series from a long tube is given below. Isthis tube acting as an open-pipe resonator or a closed-piperesonator? Exp

lain your answer. 203 Hz, 609 Hz, 1015 Hz, 1421 Hz
Physics
2 answers:
Ira Lisetskai [31]3 years ago
7 0

Answer:

We know that what we hear as a single sound or pitch when someone is speaking (for example, when making the sound [i]) is really a fundamental frequency 基頻 (determined by how many times the vocal folds vibrate in one second, and measured in cycles per second [cps], or hertz 赫 [Hz]; named after the German physicist Heinrich Rudolf Hertz), plus a whole series of harmonics 諧音 (often called 泛音 on stringed instruments) or overtones 倍音. These terms overlap in meaning, but: "harmonic" includes the fundamental frequency and all of the overtones above it, while "overtones" include all frequencies greater than the fundamental frequency.

Explanation:

    The harmonics are multiples of the fundamental frequency. So if the fundamental frequency is 100 Hz, the higher harmonics will be 200 Hz, 300 Hz, 400 Hz, 500 Hz, and so on. If the fundamental frequency were 220 Hz, the harmonics would be 440 Hz, 660 Hz, 880 Hz, and so on. In terms of intervals on the scale, we hear a base tone, its octave (eight notes up), then a note that is a twelfth up, i.e. a perfect fifth above the octave above the starting pitch, then a note two octaves up from the starting pitch, then one that is a major third above that, and on and on. If the starting pitch is middle C (C'; 256 Hz), the overtones are C" (512 Hz), G" (768 Hz): C''' (1024 Hz), E''' (1280 Hz), G''' (1536 Hz), B♭''' (1792 Hz), and so on. (Actually, your piano is tuned somewhat differently, because it uses "equal temperament". (The difference between "just intonation" and equal temperament are demonstrated in this video. But that's another story!)

   We normally don't hear the harmonics as separate tones, first of all because they have an increasingly lower amplitude than the fundamental frequency the higher up they go. The harmonics are nevertheless present in the sound, and they add a lot of richness to the sound of a human voice, a musical instrument, and many other kinds of sounds. Without them a voice would sound thin and uninteresting.

    But where do the harmonics come from, or more precisely, how are they produced? If you play the guitar, you are probably familiar with harmonics and how to produce them, even if you don't fully understand how they work. A guitar string works something like the vocal folds when it vibrates, and is a little easier to illustrate and visualize. So we will first look at how a guitar string vibrates in order to understand by analogy how the vocal folds do. Look at the animations at the bottom of this page, from the University of Salford in Manchester, UK, under Standing waves:

Pie3 years ago
7 0

Answer:

L=1.7604

Explanation:

Three successive harmonics in the pipe is given as

f_1 = 435 Hzf1=435Hz

f_2 = 725 Hzf2=725Hz

f_3 = 1015 Hzf3=1015Hz

now if we take the ratio of all frequency

then we have

f1 : f2 : f3 = 435 : 725 : 1015 = 3 : 5 : 7

since the ratio of consecutive frequency is in ratio of odd

numbers so this must be 3rd harmonics, 5th

harmonics and 7 harmonics

And as per the formula we have

f_1 = \frac{3v}{4L}f1=4L3v

435 = \frac{3(340)}{4L}435=4L3(340)

L = 1.706 mL=1.706m

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A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is rotating at 280 rev/min. It must be brought to a stop in 15.0 s.
elena-14-01-66 [18.8K]

Answer:

A) Must be done 19806.62 joules of work.

B) The average power is 1320.44 Watts.

Explanation:

A) First, we're going to use the work-energy theorem that states total work (W) done on an object is equal to the change in its kinetic energy (\Delta K):

W=\Delta K = K_{f}-K_{i} (1)

So, all we must do is to find the change on kinetic energy. Because we're working with rotational body, we should use the equation K=\frac{I\omega^{2}}{2} for the kinetic energy so:

\Delta K=\frac{I(\omega_{f})^{2}}{2}-\frac{I(\omega_{i})^{2}}{2} (2)

with \omega_{i} the initial angular velocity, \omega_{f} the final angular velocity (is zero because the wheel stops) and I the moment of inertia that for a thin hoop is I=MR^{2}, using those on (2)

\Delta K=0-\frac{MR^{2}(\omega_{i})^{2}}{2} (3)

By (3) on (1):

W= \frac{MR^{2}(\omega_{i})^{2}}{2} = \frac{(32.0)(1.2)^{2}(29.32)^{2}}{2}

W=19806.62\,J

B) Average power is work done divided by the time interval:

P=\frac{W}{\Delta t}=\frac{19806.62}{15.0}

P=1320.44\,W

NOTE: We use the relation 1rpm*\frac{2\pi}{60s}=\frac{rad}{s} to convert 280 rev/min(rpm) to 29.32 rad/s

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