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Chapter 10: Problem 70

An open pipe \(40 \mathrm{~cm}\) long and a closed pipe \(31 \mathrm{~cm}\) long, both having same diameler, are producing their first overtone, and these are in unison. Determine the end correction of these pipes.

Short Answer

Expert verified
The end correction is 4 cm.

Step by step solution

01

Understanding the Problem

An open pipe has its first overtone at its second harmonic, while a closed pipe has its first overtone at the third harmonic. Both pipes are in unison, meaning they produce the same frequency.
02

Formulate Harmonics Equations

For an open pipe, the frequency of the first overtone (second harmonic) is given by \( f_o = \frac{2v}{{2L_o + 2e}} \), where \( L_o = 0.4 \) m, and \( e \) is the end correction. For a closed pipe, the frequency of the first overtone (third harmonic) is given by \( f_c = \frac{3v}{{4L_c + 2e}} \), where \( L_c = 0.31 \) m.
03

Equate the Frequencies

Since the frequencies are equal, set the two equations for frequencies equal to each other: \[ \frac{2v}{{2L_o + 2e}} = \frac{3v}{{4L_c + 2e}} \]This implies \[ \frac{2}{2L_o + 2e} = \frac{3}{4L_c + 2e} \]
04

Solve the Equation for End Correction

Cross multiply the terms to get:\[ 2(4L_c + 2e) = 3(2L_o + 2e) \]Simplify the equation:\[ 8L_c + 4e = 6L_o + 6e \]Rearrange to find \( e \):\[ 8L_c - 6L_o = 2e \]Substitute \( L_c = 0.31 \) m and \( L_o = 0.4 \) m:\[ 8(0.31) - 6(0.4) = 2e \]Solve for \( e \):\[ 2.48 - 2.4 = 2e \]\[ 0.08 = 2e \]\[ e = 0.04 \text{ m} \] or \( e = 4 \text{ cm} \).

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Key Concepts

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

Harmonics
Harmonics in physics are the resonant frequencies at which objects naturally vibrate. These are integral multiples of a fundamental frequency, and they are extensively used in musical instruments to create various notes. For instance, when you pluck a guitar string, it not only vibrates at its fundamental frequency (the first harmonic) but also at higher frequencies called overtones.

In the context of pipes:
  • The first harmonic refers to the simplest vibration mode.
  • The second harmonic is the first overtone for an open pipe, meaning the pipe vibrates in two parts.
  • The third harmonic is the first overtone for a closed pipe, splitting into three sections.
This understanding of harmonics is crucial for solving problems involving open and closed pipes, as it helps us determine the sound frequencies produced by the pipes when they resonate.
End Correction
End correction is a subtle but crucial concept when dealing with wind instruments and pipes. It refers to the adjustment needed due to the air that slightly extends beyond the physical opening of the pipe. This extension affects the resonant frequencies and is particularly important in producing accurate sound pitches.

Mathematically, end correction (\( e \)) accounts for:
  • The extra length air travels in motion.
  • The effect of pipe end geometry on wave reflection.
For both open and closed pipes, this correction term is added to the pipe's physical length when calculating resonant frequencies. In practical problems, like the one involving pipes of different configurations, factoring in end correction ensures that calculated frequencies match expected values.
Wave Frequencies
The frequency of a wave is a measure of how often the wave passes a point in a given time frame, generally measured in hertz (Hz). It is directly related to the pitch of the sound - a higher frequency results in a higher pitch and vice versa.

In the context of pipes, different configurations lead to different resonance conditions:
  • Open pipes resonate at frequencies determined by the formula for their overtones, considering both their length and any necessary end correction.
  • Closed pipes have distinctive resonance conditions that differ from open pipes, as they resonate at odd harmonics only.
Understanding wave frequencies in pipes allows for the precise generation of musical notes, deemed essential for designing and tuning instruments like flutes and organs.
Open and Closed Pipes
Open and closed pipes are distinct in their structure and behavior, impacting their resonant frequencies and sound production:
  • Open pipes have both ends open, allowing air to move freely in and out. This leads to resonance at both even and odd harmonics. The simplest resonant condition, the first harmonic, allows the pipe to vibrate in half-wave multiples.
  • Closed pipes are closed at one end, restricting air movement, thus resonating at odd harmonics only. This means the first overtone for a closed pipe is actually the third harmonic, introducing unique sound characteristics.
These differences are vital for understanding how instruments vibrate and produce sound. In musical contexts, selecting between open and closed pipes allows for diverse pitch ranges and sound coloration, essential for creating rich and varied audio experiences.

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Most popular questions from this chapter

\(\Lambda\) tuning fork is set in vibration and the end of its handle is held firmly on a wooden table. Then (a) the loudness increases and lasis for a longer time (b) the loudness decreases, but lasts for a shorter time (c) the loudness decreases, but lasts for a longer time (d) the frequency of the sound heard cquals the original frequency of the tuning fork

Three sources of sound produce sound of the same intensity but their frequencies are 400,401 , \(402 \mathrm{~Hz}\), respectively. How many beats are heard per second?

In the state of rotation of the system as a whole there are two forces which act on the sphere: (i) elastic and (ii) centripetal. The equation of motion of the sphere is \(\frac{m d^{2} x}{d t^{2}}=-k x+m \omega^{2} x ;\) (here \(x\) sunds for displacement) or \(\frac{d^{2} x}{d t^{2}}=-\left(\frac{k}{m}-\omega^{2}\right) x\) i.e., the sphere will execute S.H.M. with $$ \begin{aligned} \omega_{n} &=\sqrt{\frac{k}{m}-\omega^{2}} \\ \therefore \quad T &=\frac{2 \pi}{\omega_{o}}=\frac{2 \pi}{\sqrt{k / m-\omega^{2}}} \\ &=\frac{2 \times 3.14}{\sqrt{\frac{20}{0.2}-(4.4 \times 4.4)}} \approx 0.7 \mathrm{~s} \end{aligned} $$ The oscillation will stop when \(T=\infty\), This is posible when $$ \omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{20}{0.2}}=10 \mathrm{rad} / \mathrm{s} $$

Which of the following waves will give rise to stationary waves? (a) \(\Lambda \sin (\omega t-k x)+A \sin (\omega t+k x)\) (b) \(A \sin (\omega t-k x)-A \sin (\omega t+k x)\) (c) \(A \sin (\omega t+k x)-A \sin (\omega t-k x)\) (d) \(A \sin ^{2}(\omega t-k x)+B \sin 3(\omega t-k x)\)

Two waves get superposed on a string \(y_{1}=3 \sin 2 \pi(x-10 t) \quad y_{2}=3 \sin 2 \pi(x+10 t)\) where \(x, y_{1}\) and \(y_{2}\) are in meter and \(t\) in sccond. Then the distance between two adjacent nodcs on the string is (a) \(25 \mathrm{~cm}\) (b) \(50 \mathrm{~cm}\) (c) \(75 \mathrm{~cm}\) (d) \(100 \mathrm{~cm}\)
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