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A musician tunes the C-string of her instrument to a fundamental frequency of \(65.4 \mathrm{~Hz}\). The vibrating portion of the string is \(0.600 \mathrm{~m}\) long and has a mass of \(14.4 \mathrm{~g}\). (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from \(65.4 \mathrm{~Hz}\) to \(73.4 \mathrm{~Hz}\), corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D}\) ?

Short Answer

Expert verified
The initial tension of the C-string must be approximately 16.54 N. To increase the frequency from 65.4 Hz to 73.4 Hz, an increase in tension of approximately 24.06% is required.

Step by step solution

01

Calculate Initial Tension

Use the formula for the frequency of a vibrating string: \(f = \sqrt{T/\mu}/2L\), where \(f\) is the frequency (65.4 Hz), \(T\) is the tension, \(\mu\) is the mass per unit length and \(L\) is the length. Rearrange the formula to solve for \(T\): \(T = \mu \cdot (2Lf)^2\). Now substitute the given values: \(\mu = 14.4/600\) g/m, \(L=0.6\) m and \(f = 65.4\) Hz. Don't forget to convert g to kg, because 1 kg = 1000 g.
02

Find the Increase in Tension

The frequency is increased to 73.4 Hz. Calculate the new tension \(T'\) using the same formula: \(T' = \mu \cdot(2L \cdot 73.4\text{Hz})^2\). Now, calculate the percentage increase in tension: Percentage = \((T'-T)/T \cdot 100%\)

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Key Concepts

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

String Tension
When it comes to things like musical instruments, string tension is a pivotal factor. It refers to how much the string is pulled tight when it is attached to the instrument. The tension affects how vibrations are produced, which in turn affects the sound. A string under higher tension will vibrate faster, producing a higher pitch sound.
Lower tension means the string vibrates more slowly, resulting in a lower pitch. One crucial equation used to calculate string tension is:
  • \( T = \mu \cdot (2Lf)^2 \)
Here, \(T\) represents tension, \(\mu\) stands for the mass per unit length, \(L\) is the length of the vibrating string, and \(f\) is the frequency.This formula shows how tension changes with frequency. The mass per unit length \(\mu\) can be calculated by dividing the total mass of the string by its length. For our example, the string is 0.6 meters long, and its mass is 14.4 grams, which needs converting to kilograms for consistency in units. This tension is what a musician needs to adjust to make sure their instrument plays the correct notes, which is essential for keeping in tune.
Fundamental Frequency
The fundamental frequency is the lowest frequency produced by a vibrating object, such as a string. This is the most basic sound the string produces and is often considered the "pitch" you hear. Instruments are typically tuned to specific fundamental frequencies to ensure they produce the exact pitches needed for a piece of music.
The fundamental frequency depends on factors like tension, length, and mass per unit length of the string.For a string under tension, the fundamental frequency \(f\) is calculated using the formula:
  • \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \)
Since it's a primary characteristic of sound, achieving the correct fundamental frequency is crucial for musical performance. The musician initially tunes the string to a fundamental frequency of 65.4 Hz, which is typical for a "C" note on the C-string.
Frequency Change Calculation
Changing the frequency of a string requires an adjustment in tension. This happens often during musical performances or tuning. Musicians need to increase or decrease the tension to change the pitch of a string, thus altering the frequency. The challenge is finding out by how much the tension needs to change for a specific frequency change.
In our case, the frequency changes from 65.4 Hz to 73.4 Hz, marking a move from a "C" note to a "D" note. This is a common practice known as tuning.To calculate the new tension \(T'\) when the frequency must change, use the same formula:
  • \( T' = \mu \cdot (2L \cdot 73.4\text{ Hz})^2 \)
The percentage change in tension is then calculated to understand how much harder or softer the string must be pulled:
  • Percentage Increase = \( \frac{T' - T}{T} \cdot 100\% \)
This calculation helps musicians know exactly how much they need to adjust their instruments in terms of tension. Ensuring precise tension changes means better sound quality and accurate pitches.

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Most popular questions from this chapter

A transverse sine wave with an amplitude of \(2.50 \mathrm{~mm}\) and a wavelength of \(1.80 \mathrm{~m}\) travels from left to right along a long, horizontal, stretched string with a speed of \(36.0 \mathrm{~m} / \mathrm{s}\). Take the origin at the left end of the undisturbed string. At time \(t=0\) the left end of the string has its maximum upward displacement. (a) What are the frequency, angular frequency, and wave number of the wave? (b) What is the function \(y(x, t)\) that describes the wave? (c) What is \(y(t)\) for a particle at the left end of the string? (d) What is \(y(t)\) for a particle \(1.35 \mathrm{~m}\) to the right of the origin? (e) What is the maximum magnitude of transverse velocity of any particle of the string? (f) Find the transverse displacement and the transverse velocity of a particle \(1.35 \mathrm{~m}\) to the right of the origin at time \(t=0.0625 \mathrm{~s}\)

Standing waves are produced on a string that is held fixed at both ends. The tension in the string is kept constant. (a) For the second overtone standing wave the node-to-node distance is \(8.00 \mathrm{~cm} .\) What is the length of the string? (b) What is the node-to-node distance for the fourth harmonic standing wave?

For a violin, estimate the length of the portions of the strings that are free to vibrate. (a) The frequency of the note played by the open E5 string vibrating in its fundamental standing wave is 659 Hz. Use your estimate of the length to calculate the wave speed for the transverse waves on the string. (b) The vibrating string produces sound waves in air with the same frequency as that of the string. Use \(344 \mathrm{~m} / \mathrm{s}\) for the speed of sound in air and calculate the wavelength of the E5 note in air. Which is larger: the wavelength on the string or the wavelength in air? (c) Repeat parts (a) and (b) for a bass viol, which is typically played by a person standing up. Start your calculation by estimating the length of the bass viol string that is free to vibrate. The G2 string produces a note with frequency \(98 \mathrm{~Hz}\) when vibrating in its fundamental standing wave.

A thin, taut string tied at both ends and oscillating in its third harmonic has its shape described by the equation \(y(x, t)=(5.60 \mathrm{~cm}) \sin [(0.0340 \mathrm{rad} / \mathrm{cm}) x] \sin [(50.0 \mathrm{rad} / \mathrm{s}) t],\) where the origin is at the left end of the string, the \(x\) -axis is along the string, and the \(y\) -axis is perpendicular to the string. (a) Draw a sketch that shows the standing-wave pattern. (b) Find the amplitude of the two traveling waves that make up this standing wave. (c) What is the length of the string? (d) Find the wavelength, frequency, period, and speed of the traveling waves. (e) Find the maximum transverse speed of a point on the string. (f) What would be the equation \(y(x, t)\) for this string if it were vibrating in its eighth harmonic?

A wire with mass \(40.0 \mathrm{~g}\) is stretched so that its ends are tied down at points \(80.0 \mathrm{~cm}\) apart. The wire vibrates in its fundamental mode with frequency \(60.0 \mathrm{~Hz}\) and with an amplitude at the antinodes of \(0.300 \mathrm{~cm}\). (a) What is the speed of propagation of transverse waves in the wire? (b) Compute the tension in the wire. (c) Find the maximum transverse velocity and acceleration of particles in the wire.

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