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Two strings are adjusted to vibrate at exactly \(200 \mathrm{Hz}\). Then the tension in one string is increased slightly. Afterward, three beats per second are heard when the strings vibrate at the same time. What is the new frequency of the string that was tightened?

Short Answer

Expert verified
The new frequency of the string that was tightened is 203 Hz.

Step by step solution

01

Understanding Beats

When two waves of different frequencies interfere, the resulting pattern is characterized by points where the amplitude of the wave is momentarily larger or smaller than the amplitude of each of the component waves. This is known as beating. The beat frequency is equal to the absolute difference in frequencies of the two waves.
02

Calculating Change in Frequency

In this case, it is given that 3 beats per second are heard. This means that the difference in frequency of the two strings is 3 Hz.
03

Finding New Frequency

Given that the original frequency of both strings was 200 Hz, and knowing that the beat frequency represents the difference in frequencies, the new frequency of the tightened string can be calculated as the original frequency plus the beat difference. Therefore, adding 3 Hz to the original frequency of 200 Hz, we find that the new frequency is 203 Hz.

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Key Concepts

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

Wave Interference
Wave interference happens when two or more waves meet while traveling through the same medium. It’s like when you throw two stones into a pond and see the ripples collide.

This interaction can either reinforce the wave, making it bigger, or cancel part of it out, making it smaller.
  • **Constructive interference** occurs if the waves align perfectly, leading to amplification, such as louder sounds in acoustics.
  • **Destructive interference** happens when the waves do not align, canceling each other out partially or completely, leading to softer sounds.
This principle is what causes beats when sound waves of slightly different frequencies interfere. You’ll hear the sound getting louder and softer at regular intervals, which is the beat frequency.

Frequency Difference
The frequency difference is the key to understanding beats. When two sound waves with slightly different frequencies interfere, they create a third wave with a frequency that is the difference between the two original frequencies.

To understand this better, picture two waves occurring slightly out of sync:
  • **Beat frequency** is simply the absolute value of the difference between these two frequencies.
  • For example, if you hear 3 beats per second, the frequency difference is 3 Hz.
In the given problem, the original frequency is 200 Hz, and the beat frequency is 3 Hz. So, the frequencies differ by 3 Hz, leading to the new frequency being either 197 Hz or 203 Hz. Since one string was tightened, resulting in a higher frequency, the answer is 203 Hz.

Musical Acoustics
Musical acoustics is the science of sound as it relates to music. In musical acoustics, understanding how different frequencies interact is crucial.

Here, the concepts of beats and frequency differences directly apply. When two notes are played together, their frequencies can create beats, giving rise to harmonious or dissonant sounds depending on the frequencies' ratios.
  • **Harmony** results when frequencies are integer multiples or small whole number ratios.
  • **Dissonance** occurs with awkward ratios, frequently producing noticeable beats.
Musicians and instrument tuners use this principle to ensure that instruments are in tune. By listening to the rate and intensity of beats produced by interacting sound waves, they can finely tune their instruments. In the given problem, the resulting 3 Hz beat indicates a subtle but perceptible tension change in the string.

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Most popular questions from this chapter

The speed of sound in room temperature \(\left(20^{\circ} \mathrm{C}\right)\) air is \(343 \mathrm{m} / \mathrm{s} ;\) in room temperature helium, it is \(1010 \mathrm{m} / \mathrm{s} .\) The fundamental frequency of an open-closed tube is 315 Hz when the tube is filled with air. What is the fundamental frequency if the air is replaced with helium?

A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of \(39.1 \mathrm{cm} .\) The room is slightly warm, so the speed of sound is \(344 \mathrm{m} / \mathrm{s}\). If the string's linear density is \(0.600 \mathrm{g} / \mathrm{m}\) and the tension is \(150 \mathrm{N},\) how long is the vibrating section of the violin string?

Two identical loudspeakers separated by distance \(d\) emit \(170 \mathrm{Hz}\) sound waves along the \(x\) -axis. As you walk along the axis, away from the speakers, you don't hear anything even though both speakers are on. What are three possible values for d? Assume a sound speed of \(340 \mathrm{m} / \mathrm{s}\)

What are the three longest wavelengths for standing waves on a 240 -cm-long string that is fixed at both ends? b. If the frequency of the second-longest wavelength is \(50.0 \mathrm{Hz},\) what is the frequency of the third-longest wavelength?

A \(40-\mathrm{cm}\) -long tube has a \(40-\mathrm{cm}-\) long insert that can be pulled in and out, as shown in Figure \(\mathrm{P} 16.61 .\) A vibrating tuning fork is held next to the tube. As the insert is slowly pulled out, the sound from the tuning fork creates standing waves in the tube when the total length \(L\) is \(42.5 \mathrm{cm}, 56.7 \mathrm{cm},\) and \(70.9 \mathrm{cm} .\) What is the frequency of the tuning fork? The air temperature is \(20^{\circ} \mathrm{C}\).

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