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A person is playing a small fute 10.75 \(\mathrm{cm}\) long, open at one end and closed at the other, near a taut string having a fundamental frequency of 600.0 \(\mathrm{Hz}\) . If the speed of sound is 344.0 \(\mathrm{m} / \mathrm{s}\) , for which harmonics of the flute will the string resonate? In each case, which harmonic of the string is in resonance?

Short Answer

Expert verified
The string resonates with the 3rd harmonic of the flute, which corresponds to the string's 4th harmonic.

Step by step solution

01

Understand the Problem

We have a flute which is closed at one end and open at the other and a taut string with a fundamental frequency of 600 Hz. We need to find the harmonics of the flute that cause the string to resonate, considering the speed of sound in air is 344 m/s.
02

Calculate Fundamental Frequency of Flute

The formula for the fundamental frequency of a pipe closed at one end is \( f_1 = \frac{v}{4L} \), where \(v\) is the speed of sound and \(L\) is the length of the flute. Substitute \(v = 344 \, \mathrm{m/s}\) and \(L = 0.1075 \, \mathrm{m}\).\[ f_1 = \frac{344}{4 \times 0.1075} = 800 \, \mathrm{Hz}\]
03

Determine Flute Harmonics Frequencies

The harmonics in a closed pipe are odd multiples of the fundamental frequency: \( f_n = (2n-1)f_1 \) for \(n = 1, 2, 3, \ldots\) Calculate the first few harmonics:- 1st harmonic: \( f_1 = 800 \, \mathrm{Hz}\)- 3rd harmonic: \( f_3 = 2400 \, \mathrm{Hz}\)- 5th harmonic: \( f_5 = 4000 \, \mathrm{Hz}\)
04

Match Flute Harmonics to String Harmonics

The string has a fundamental frequency of 600 Hz and will resonate at its harmonics: 600, 1200, 1800, 2400, etc., which are integer multiples of the fundamental frequency. Compare these with the flute's harmonics: - Flute's 1st harmonic (800 Hz): No resonance with string. - Flute's 3rd harmonic (2400 Hz): Resonates with the string's 4th harmonic (2400 Hz). - Flute's 5th harmonic (4000 Hz): No corresponding resonance.
05

Final Answer

The string resonates with the 3rd harmonic of the flute. This corresponds to the 4th harmonic of the string since \(2400 \, \mathrm{Hz}\) is the common frequency where both systems resonate.

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Key Concepts

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

Fundamental Frequency
The fundamental frequency is a crucial concept in acoustics. It's the lowest frequency at which a system resonates. Imagine a flute or string as a playground for sound waves. The fundamental frequency is like the main swing on this playground, setting the tone for all the other swings, or harmonics, around it.
For a taut string, the fundamental frequency is given directly. In this exercise, it's 600 Hz. This means the string naturally vibrates at 600 oscillations per second.
The fundamental frequency in a closed pipe, like our flute, is calculated using the formula:
  • \(f_1 = \frac{v}{4L}\)
Here, \(v = 344\,\text{m/s}\) is the speed of sound, and \(L = 0.1075\,\text{m}\) is the length of the flute. This calculation gives us a fundamental frequency of 800 Hz for the flute. This is where our flute naturally wants to resonate.
Speed of Sound
The speed of sound is an essential factor in determining how sound travels. It's the rate at which sound waves move through a medium, like air. For our scenario, the speed of sound is 344 m/s. This is a typical value for air at room temperature.
The speed of sound affects how frequency translates to physical length in instruments. For instance, in our flute, it helps determine the fundamental frequency by impacting how long the wave takes to bounce back and forth inside the flute.
In practical terms, knowing the speed of sound allows us to bridge the gap between the physical dimensions of musical instruments and the frequencies they produce. It's like understanding a road's speed limit when planning a trip. Sound waves travel at this 'limit,' interpreting physical dimensions into musical notes.
Resonance
Resonance is the magic that happens when one system vibrates at the same frequency as another system. Picture two tuning forks near each other. Striking one causes the other to vibrate without being touched, thanks to resonance.
In our exercise, resonance occurs when the frequencies of the flute and the string align. This can happen at harmoni c frequencies, which are multiples of the fundamental frequencies.
When the flute's 3rd harmonic at 2400 Hz corresponds to the string's 4th harmonic, both systems resonate. It's like finding a groove in music where different instruments sync perfectly. Resonance amplifies sound, making it richer and more vibrant, which is why a flute can influence the sound a string emits at matching harmonics.
Closed Pipe Harmonics
Understanding closed pipe harmonics is key to knowing how certain musical instruments work. In a closed pipe, like our flute, only odd numbered harmonics are possible. This means that the resonant frequencies occur at odd multiples of the fundamental frequency.
The harmonics are given by the formula:
  • \(f_n = (2n-1)f_1\)
where \(n\) is an integer representing the harmonic number.
In our flute:
  • 1st harmonic is 800 Hz.
  • 3rd harmonic is 2400 Hz.
  • 5th harmonic is 4000 Hz.
Closed pipe harmonics explain why only certain higher pitches are heard when you play such an instrument. The absence of even harmonics gives a distinctive sound, making instruments like the bassoon and oboe sound unique. These harmonics also provide opportunities for resonance with other instruments or systems, as explored in our exercise.

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Most popular questions from this chapter

Longitudinal Waves in Different Fluids. (a) A longitudinal wave propagating in a water-filled pipe has intensity \(3.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\) and frequency 3400 \(\mathrm{Hz}\) . Find the amplitude \(A\) and wavelength \(\lambda\) of the wave. Water has density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) and bulk modulus \(2.18 \times 10^{9} \mathrm{Pa}\) . Water has density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) at pressure \(1.00 \times 10^{5} \mathrm{Pa}\) and density 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) , what will be the amplitude \(A\) and wavelength \(\lambda\) of a longitudinal wave with the same intensity and frequency as in part (a)? (c) In which fluid is the amplitude larger, water or air? What is the ratio of the two amplitudes? Why is this ratio so different from 1.00\(?\)

A standing wave with a frequency of 1100 \(\mathrm{Hz}\) in a column of methane \(\left(\mathrm{CH}_{4}\right)\) at \(20.0^{\circ} \mathrm{C}\) produces nodes that are 0.200 \(\mathrm{m}\) apart. What is the value of \(\gamma\) for methane? (The molar mass of methane is 16.0 \(\mathrm{g} / \mathrm{mol}\) )

(a) In a liquid with density 1300 \(\mathrm{kg} / \mathrm{m}^{3}\) , longitudinal waves with frequency 400 \(\mathrm{Hz}\) are found to have wavelength 8.00 \(\mathrm{m}\) . Calculate the bulk modulus of the liquid. (b) A metal bar with a length of 1.50 \(\mathrm{m}\) has density 6400 \(\mathrm{kg} / \mathrm{m}^{3}\) . Longitudinal sound waves take \(3.90 \times 10^{-4} \mathrm{s}\) to travel from one end of the bar to the other. What is Young's modulus for this metal?

An oscillator vibrating at 1250 Hz produces a sound wave that travels through an ideal gas at 325 \(\mathrm{m} / \mathrm{s}\) when the gas temperature is \(22.0^{\circ} \mathrm{C}\) . For a certain experiment, you need to have the same uscillator produce sound of waveleugth 28.5 \(\mathrm{cm}\) in this gas. What should the gas temperature be to achieve this wavelength?

Example 16.1 (Section 16.1\()\) showed that for sound waves in air with frequency 1000 \(\mathrm{Hz}\) , a displacement amplitude of\( 1.2 \times 10^{-8} \mathrm{m}\) produces a pressure amplitude of \(3.0 \times 10^{-2} \mathrm{Pa}\) . Water at \(20^{\circ} \mathrm{C}\) has a bulk modulus of \(2.2 \times 10^{9} \mathrm{Pa}\) , and the speed of sound in water at this temperature is 1480 \(\mathrm{m} / \mathrm{s}\) . For \(1000-\mathrm{Hz}\) sound waves in \(20^{\circ} \mathrm{C}\) water, what displacement amplitude is produced if the pressure amplitude is \(3.0 \times 10^{-2}\) Pa? Explain why your answer is much less than \(1.2 \times 10^{-8} \mathrm{Pa}\) .

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