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One of the \(63.5-\mathrm{cm}\) -long strings of an ordinary guitar is tuned to produce the note \(\mathrm{B}_{3}\) (frequency \(245 \mathrm{~Hz}\) ) when vibrating in its fundamental mode. (a) Find the speed of transverse waves on this string. (b) If the tension in this string is increased by \(1.0 \%,\) what will be the new fundamental frequency of the string? (c) If the speed of sound in the surrounding air is \(344 \mathrm{~m} / \mathrm{s},\) find the frequency and wavelength of the sound wave produced in the air by the vibration of the \(\mathrm{B}_{3}\) string. How do these compare to the frequency and wavelength of the standing wave on the string?

Short Answer

Expert verified
Wave speed: 311.15 m/s; New frequency: 246.15 Hz; Air frequency: 245 Hz; Air wavelength: 1.404 m.

Step by step solution

01

Identify Given Data and Formula for Wave Speed

We are given the string length, \( L = 63.5 \text{ cm} = 0.635 \text{ m} \), and the fundamental frequency, \( f = 245 \text{ Hz} \). In the fundamental mode, the wavelength \( \lambda \) of the standing wave is twice the length of the string, so \( \lambda = 2L = 1.270 \text{ m} \). The wave speed \( v \) on the string is given by the formula \( v = f \cdot \lambda \).
02

Calculate Speed of Transverse Waves

Using the formula from Step 1, we calculate the wave speed: \[v = 245 \text{ Hz} \times 1.270 \text{ m} = 311.15 \text{ m/s}\]
03

Adjust Tension and Find New Frequency

If the tension is increased by \(1\%\), the wave speed \( v' \) changes according to the square root of the tension, so \( v' = v \times \sqrt{1.01} \). Calculate \( v' \) as follows:\[v' = 311.15 \text{ m/s} \times \sqrt{1.01} = 312.72 \text{ m/s}\]The new fundamental frequency \( f' \) is:\[f' = \frac{v'}{\lambda} = \frac{312.72 \text{ m/s}}{1.270 \text{ m}} = 246.15 \text{ Hz}\]
04

Calculate Sound Frequency and Wavelength in Air

The frequency of the sound wave in air is equal to the frequency of vibration of the string, which is \( 245 \text{ Hz} \). The speed of sound in air is given as \( 344 \text{ m/s} \). The wavelength \( \lambda_{air} \) of the sound wave in air is:\[\lambda_{air} = \frac{344 \text{ m/s}}{245 \text{ Hz}} = 1.404 \text{ m}\]
05

Compare Frequencies and Wavelengths

The frequencies of the sound wave in air (\( 245 \text{ Hz} \)) and the string’s vibration are identical. However, the wavelength of the sound wave in air (\( 1.404 \text{ m} \)) is different from the wavelength of the standing wave on the string (\( 1.270 \text{ m} \)).

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Key Concepts

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

Transverse Waves
Transverse waves are a type of wave where the oscillations occur perpendicular to the direction of the wave's travel. In other words, if the wave is moving horizontally, the vibrations are vertical. Transverse waves can be found in various physical phenomena, including light and mechanical waves on a string.

Sound waves require a medium to propagate, such as air or water. However, it's essential to understand that sound waves themselves are typically longitudinal, not transverse, in nature. On a string, however, the waves generated by plucking or striking the string manifest as transverse waves.

The speed of transverse waves on a string depends on two main factors:
  • The tension in the string.
  • The mass per unit length of the string.
As the tension increases, the wave speed also increases because the string becomes taut, enabling faster propagation of waves. Conversely, a heavier string will have a slower wave speed due to its greater inertia.
String Instruments Physics
String instruments, such as guitars, violins, and pianos, produce sound primarily through the vibration of strings. When a string is plucked or bowed, it vibrates and creates transverse waves that travel along its length. These vibrations are fundamental to creating the instruments' musical tones.

The pitch of the sound produced by a stringed instrument is controlled by three main factors:
  • The tension of the string.
  • The mass per unit length of the string.
  • The length of the string.
By adjusting these factors—tightening or loosening the strings, for instance—musicians can tune their instruments to achieve the desired pitch.

In the practical context, when you increase the tension, you also increase the wave speed and frequency, leading to a higher-pitched sound. This principle is at play in the exercise, where adjusting the guitar string's tension changes the fundamental frequency.
Fundamental Frequency
The fundamental frequency, also known as the first harmonic, is the lowest frequency at which a string vibrates when excited. It corresponds to the simplest mode of vibration, where the string has only one antinode at the center and nodes at the two ends.

In practice, when a guitar string is plucked, it predominantly vibrates at its fundamental frequency, which defines the note that you hear. The fundamental frequency can be calculated using the formula:\[ f_1 = \frac{v}{2L} \]where:
  • \( f_1 \) is the fundamental frequency,
  • \( v \) is the speed of the wave on the string,
  • \( L \) is the length of the string.
Understanding the fundamental frequency is crucial for tuning instruments, as this base frequency determines the harmonic series of the notes below and above it.
Speed of Sound
The speed of sound refers to how fast sound waves travel through a medium, such as air, water, or solids. At room temperature, the speed of sound in the air is approximately 344 meters per second. This speed can vary slightly based on atmospheric conditions like temperature, pressure, and humidity.

While sound waves on a string are transverse, the actual sound waves you hear in the air are longitudinal. They consist of compressions and rarefactions of air molecules, which your ear interprets as sound.

In the example given, we explored not only the vibration of the guitar string but also the sound wave it generates in the air. Interestingly, the frequency of the sound wave in the air remains the same as that of the string's vibration; however, the wavelengths differ due to the different wave speeds in the medium (string vs. air). Understanding both components helps musicians, acousticians, and physics students appreciate and manage sound production and transmission.

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Most popular questions from this chapter

(a) If the amplitude in a sound wave is tripled, by what factor does the intensity of the wave increase? (b) By what factor must the amplitude of a sound wave be decreased in order to decrease the intensity by a factor of \(3 ?\)

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A trumpet player is tuning his instrument by playing an A note simultaneously with the first-chair trumpeter, who has perfect pitch. The first-chair player's note is exactly \(440 \mathrm{~Hz}\), and 2.8 beats per second are heard. What are the two possible frequencies of the other player's note?

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