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When the string of a guitar is pressed against a fret, the shortened string vibrates at a frequency \(5.95 \%\) higher than when the previous fret is pressed. If the length of the part of the string that is free to vibrate is \(64.8 \mathrm{cm},\) how far from one end of the string are the first three frets located?

Short Answer

Expert verified
Answer: The first three frets on the string are located 3.862 cm, 7.5212 cm, and 10.9715 cm away from one end of the string, respectively.

Step by step solution

01

Understand the relationship between frequency and length

According to the frequency formula for a vibrating string, we can say that the frequency (f) is inversely proportional to the length (L) of the string: \(f \propto \frac{1}{L}\) So, when the length decreases, the frequency increases.
02

Calculate the length factor

In this case, the frequency increases by \(5.95 \%\) from one fret to another. Since the frequency is inversely proportional to the length, we can say that the decrease in length is also \(5.95 \%\). We can express it as a length factor: Length factor = \(1 - \frac{5.95}{100} = 1 - 0.0595 = 0.9405\) As the length factor is less than 1, it implies the length of the vibrating part decreases as we move from one fret to another.
03

Calculate the length of the vibrating string for each fret

As we know the initial length of the vibrating string (64.8 cm) and the length factor, we can calculate the length for each of the first three frets: Fret 1 length: \(64.8 \cdot 0.9405 = 60.938 \mathrm{cm}\) Fret 2 length: \(60.938 \cdot 0.9405 = 57.2788 \mathrm{cm}\) Fret 3 length: \(57.2788 \cdot 0.9405 = 53.8285 \mathrm{cm}\)
04

Find the distance between the vibrating string and the frets

Finally, we want to find the distances between the frets and one end of the string. We will calculate these distances by subtracting the length of the vibrating string of each fret from the original string length: Fret 1 distance: \(64.8 \mathrm{cm} - 60.938 \mathrm{cm} = 3.862 \mathrm{cm}\) Fret 2 distance: \(64.8 \mathrm{cm} - 57.2788 \mathrm{cm} = 7.5212 \mathrm{cm}\) Fret 3 distance: \(64.8 \mathrm{cm} - 53.8285 \mathrm{cm} = 10.9715 \mathrm{cm}\) So, the first three frets are located \(3.862 \mathrm{cm}\), \(7.5212 \mathrm{cm}\), and \(10.9715 \mathrm{cm}\) away from one end of the string, respectively.

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Most popular questions from this chapter

(a) Use a graphing calculator or computer graphing program to plot \(y\) versus \(x\) for the function $$ y(x, t)=(5.0 \mathrm{cm})[\sin (k x-\omega t)+\sin (k x+\omega t)] $$ for the times \(t=0,1.0 \mathrm{s},\) and \(2.0 \mathrm{s}\). Use the values \(k=\pi /(5.0 \mathrm{cm})\) and \(\omega=(\pi / 6.0) \mathrm{rad} / \mathrm{s} .\) (b) Is this a traveling wave? If not, what kind of wave is it?

A sound wave with intensity \(25 \mathrm{mW} / \mathrm{m}^{2}\) interferes constructively with a sound wave that has an intensity of $15 \mathrm{mW} / \mathrm{m}^{2} .$ What is the intensity of the superposition of the two? (tutorial: superposition)
The equation of a wave is $$ y(x, t)=(3.5 \mathrm{cm}) \sin \left\\{\frac{\pi}{3.0 \mathrm{cm}}[x-(66 \mathrm{cm} / \mathrm{s}) t]\right\\} $$ Find (a) the amplitude and (b) the wavelength of this wave.
A string \(2.0 \mathrm{m}\) long is held fixed at both ends. If a sharp blow is applied to the string at its center, it takes \(0.050 \mathrm{s}\) for the pulse to travel to the ends of the string and return to the middle. What is the fundamental frequency of oscillation for this string?
A harpsichord string of length \(1.50 \mathrm{m}\) and linear mass density $25.0 \mathrm{mg} / \mathrm{m}\( vibrates at a (fundamental) frequency of \)450.0 \mathrm{Hz}$. (a) What is the speed of the transverse string waves? (b) What is the tension? (c) What are the wavelength and frequency of the sound wave in air produced by vibration of the string? (The speed of sound in air at room temperature is \(340 \mathrm{m} / \mathrm{s} .\) )
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