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The portion of the string of a certain musical instrument between the bridge and upper end of the finger board (that part of the string that is free to vibrate) is 60.0 \(\mathrm{cm}\) long. and this length of the string has mass 2.00 g. The string sounds an \(\mathrm{A}_{4}\) note \((440 \mathrm{Hz})\) when played. (a) Where must the player put a finger (what distance \(x\) from bridge) to play \(\mathbf{a}\) Ds note \((587 \quad \mathrm{Hz}) ? \quad\) (See Fig. 15.36 . For both the A_{4} and \(D_{5}\) notes, the string vibrates in its fundamental mode. (b) Without retuning, is it possible to play a G \(_{4}\) note \((392 \mathrm{Hz})\) on this string? Why or why not?

Short Answer

Expert verified
(a) Place the finger 15.02 cm from the bridge. (b) No, playing G_4 is not possible without retuning.

Step by step solution

01

Understanding the Frequency-Pitch Relationship

To determine where the finger should be placed to change the pitch from A_4 (440 Hz) to D_5 (587 Hz), we need to understand that the frequency of a vibrating string is proportional to the inverse of its vibrating length. This relationship is described by the formula: \[ f \propto \frac{1}{L} \] where \( f \) is the frequency and \( L \) is the length of the vibrating part of the string.
02

Setting Up the Proportion

Given the initial length \( L_0 = 60.0 \) cm for A_4 (440 Hz), and the required frequency for D_5 is 587 Hz, we set up the proportional relationship: \[ \frac{440}{587} = \frac{x}{60} \] where \( x \) is the new vibrating length needed to produce 587 Hz.
03

Solving the Proportion

To find \( x \), solve the equation obtained in Step 2: \[ x = 60 \times \frac{440}{587} = 44.98 \text{ cm} \] This means the player's finger should be placed 60 - 44.98 = 15.02 cm from the bridge to achieve the D_5 note.
04

Determining the Possibility of Playing G_4

To check if G_4 (392 Hz) can be played, consider the proportion \( \frac{440}{392} \): \[ \frac{440}{392} \approx 1.12 \] which suggests that the vibrating length must be proportionally longer than for A_4, or approximately 67.68 cm, which exceeds the original length of 60 cm. This is not possible on this fixed string without lengthening.

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Key Concepts

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

Frequency-Pitch Relationship
The frequency of a musical note is directly related to its pitch: a high frequency corresponds to a high pitch, and a low frequency corresponds to a low pitch. On a stringed instrument, this relationship between frequency and pitch is crucial for understanding how different notes are produced.

When a string on an instrument vibrates, it creates sound waves at a certain frequency. This frequency is determined by how fast the string vibrates, which is influenced by factors like tension, mass, and length of the string.

For a fixed string with a constant tension and uniform material, the frequency is inversely proportional to the length according to the formula:\[ f \propto \frac{1}{L} \]where \( f \) represents frequency, and \( L \) is the vibrating length of the string.

This means that shortening the string increases the frequency, resulting in a higher pitch, and lengthening the string decreases the frequency, producing a lower pitch.
Vibrating String Length
The length of the string that is free to vibrate is a critical factor in determining the frequency of the note produced. When a musician places a finger on the string, they change its effective vibrating length.

The remaining part of the string acts as the new length that determines the new frequency. For example, to play different notes on a violin, the player presses down on various points along the string, effectively shortening or lengthening it to achieve the desired pitch.

The basic principle involves calculating where the finger should be placed to reach the desired pitch. This calculation uses the known relationship between frequency and length, allowing us to determine the precise point along the string needed to create a specific note. For instance, going from an \( A_4 \) (440 Hz) to a \( D_5 \) (587 Hz) note involves using the equation:\[ \frac{f_1}{f_2} = \frac{L_2}{L_1}\]Where \( f_1 \) and \( f_2 \) are the initial and final frequencies, and \( L_1 \) and \( L_2 \) are the initial and new lengths of the string.
String Instrument Tuning
Tuning a string instrument involves adjusting the tension or length of the strings so they produce the correct pitch for a given note. Musicians often use tuning pegs to modify the tension of the string, which is one effective way to change the pitch without altering the vibrating length.

However, on instruments like violins or guitars, players can also adjust the pitch in real time during performances by placing fingers at different positions on the strings. This action changes the vibrating length, thus altering the frequency of the sound waves produced.

Besides tuning by length or tension, other factors like temperature and humidity can also impact string pitch. These factors can cause changes in string tension, which musicians must frequently adjust for optimal sound.
Harmonic Modes of Vibration
Strings on musical instruments not only vibrate at their fundamental frequency but also at additional frequencies called harmonics or overtones. These represent multiple ways in which the string can vibrate simultaneously, and they enrich the sound produced by adding complexity and depth.

Each harmonic corresponds to a unique mode of vibration. The fundamental mode is the simplest, involving the whole string vibrating as one segment. More complex modes involve the string vibrating in segments, such as halves, thirds, fourths, etc., each corresponding to higher harmonic frequencies.

For example, when an open string is played, it predominantly vibrates in its fundamental mode, producing a clean note. When a player lightly touches the string at certain points, these divide the string and produce distinct harmonics. Such techniques are often employed to enrich the musical sounds and textures available to musicians.

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Most popular questions from this chapter

Three pieces of string, each of length \(L,\) are joined together end to end, to make a combined string of length 3\(L\) . The first piece of string has mass per unit length \(\mu_{1}\) , the second piece has mass per unit length \(\mu_{2}=4 \mu_{1},\) and the third piece has mass per unit length \(\mu_{3}=\mu_{1} / 4\) . (a) If the combined string is under tension \(F\) ,how much time does it take a transverse wave to travel the entire length 3\(L ?\) Give your answer in terms of \(L, F,\) and \(\mu_{1}\) . (b) Does your answer to part (a) depend on the order in which the three pieces are joined together? Explain.

Visible Light. Light is a wave, bnt not a mechanical wave. The quantities that oscillate are electric and magnetic fields. Light visible to humans has wavelengths between 400 \(\mathrm{nm}\) (violet) and \(700 \mathrm{nm}(\mathrm{red}),\) and all light travels through vacuum at speed \(c=3.00 \times 10^{8} \mathrm{m} / \mathrm{s}\) (a) What are the limits of the frequency and period of visible light? (b) Could you time a single light vibration with a stopwatch?

Guitar String. One of the 63.5 -cm-long strings of an ordinary guitar is tuned to produce the note \(\mathbf{B}_{3}\) (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?

A string with both ends held fixed is vibrating in its third harmonic. The waves have a speed of 192 \(\mathrm{m} / \mathrm{s}\) and a frequency of 240 \(\mathrm{Hz}\) . The amplitude of the standing wave at an antinode is 0.400 \(\mathrm{cm}\) . (a) Calculate the amplitude at points on the string a distance of (i) \(40.0 \mathrm{cm} ;\) (ii) \(20.0 \mathrm{cm} ;\) and (iii) 10.0 \(\mathrm{cm}\) from the left end of the string. (b) At each point in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement? (c) Calculate the maximum transverse velocity and the maximum transverse acceleration of the string at each of the points in part (a).

Waves of Arbitrary Shape. (a) Explain why any wave described by a function of the form \(y(x, t)=f(x-v t)\) moves in the \(+x\) -direction with speed \(v\) . (b) Show that \(y(x, t)=f(x-v t)\) satisfies the wave equation, no matter what the functional form of \(f\) . To do this, write \(y(x, t)=f(u),\) where \(u=x-v t\) . Then, to take partial derivatives of \(y(x, t),\) use the chain rule: $$ \begin{array}{l}{\frac{\partial y(x, t)}{\partial t}=\frac{d f(u)}{d u} \frac{\partial u}{\partial t}=\frac{d f(u)}{d u}(-v)} \\ {\frac{\partial y(x, t)}{\partial x}=\frac{d f(u)}{d u} \frac{\partial u}{\partial x}=\frac{d f(u)}{d u}}\end{array} $$ (c) A wave pulse is described by the function \(y(x, t)=D e^{-(B x-C)^{2}}\) where \(B, C,\) and \(D\) are all positive constants. What is the speed of this wave?

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