91Ó°ÊÓ

You are designing a two-string instrument with metal strings \(35.0 \mathrm{~cm}\) long, as shown in Fig. \(\mathrm{P} 15.52 .\) Both strings are under the same tension. String \(S_{1}\) has a mass of \(8.00 \mathrm{~g}\) and produces the note middle \(\mathrm{C}\) (frequency \(262 \mathrm{~Hz}\) ) in its fundamental mode. (a) What should be the tension in the string? (b) What should be the mass of string \(S_{2}\) so that it will produce A-sharp (frequency \(466 \mathrm{~Hz}\) ) as its fundamental? (c) To extend the range of your instrument, you include a fret located just under the strings but not normally touching them. How far from the upper end should you put this fret so that when you press \(S_{1}\) tightly against it, this string will produce \(\mathrm{C}\) -sharp (frequency \(277 \mathrm{~Hz}\) ) in its fundamental? That is, what is \(x\) in the figure? (d) If you press \(S_{2}\) against the fret, what frequency of sound will it produce in its fundamental?

Step by step solution

01

Calculate the tension in string S1

Use the formula \(f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\) to compute the tension \(T\). First, calculate the linear density \(\mu\) of the string, using the given mass and length. \(\mu = \frac{mass}{length} = \frac{8.00g}{35.0cm} = 0.22857 g/cm = 228.57 g/m\). Now, rearrange the formula to solve for \(T\): \(T = (f * 2L)^2 * \mu\).Substitute the known values: \(T = (262Hz * 2*0.35m)^2 * 228.57 g/m = 79.69 N\).

02

Calculate the mass of string S2

The problem states that string S2 is under the same tension and same length as string S1, but produces a different fundamental frequency. Thus, its linear density must differ.First, rearrange the formula to solve for \(\mu\) this time: \(\mu = \frac{(f * 2L)^2}{T}\).Substitute the known values for \(f = 466 Hz\), \(L = 0.35 m\), \(T = 79.69 N\). We find \(\mu = 0.41736 kg/m\).Now, convert \(\mu\) back to mass by multiplying with length: \(mass = \mu * length = 0.41736 kg/m * 0.35 m = 0.015 kg = 15g\).

03

Calculate the location of the fret for string S1 to produce C-sharp

Pressing the fret on string S1 changes the effective vibrating length of the string. We need to find this new length (\(L'\)) for when the string produces the note C-sharp (frequency 277 Hz). Use the formula for fundamental frequency, but this time solve for \(L'\): \(L' = \frac{1}{2f}\sqrt{\frac{T}{\mu}}\).Substitute the known values for \(T\), \(f\), and \(\mu\) and solve to find \(L' = 0.318 m\).Now, the location of the fret (\(x\)) is given by \(x = L - L'\) = 0.35m - 0.318m = 0.032m = 3.2cm.

04

Calculate the new frequency produced by string S2 when pressed against the fret

The problem asks for the new fundamental frequency when string S2 is pressed against the fret at location \(x\). Since the effective length of string S2 is now also \(L'\), we can again use \(f = \frac{1}{2L'}\sqrt{\frac{T}{\mu}}\) to solve for the new frequency.Substitute the known values: \(f' = \frac{1}{2*0.318m}\sqrt{\frac{79.69N}{0.41736 kg/m}} = 501 Hz.\)

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Key Concepts

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

Tension in Strings

The tension in a string has a direct influence on the frequency of the sound wave that the string produces when it is vibrated. Tension is the force stretching the string, and it's measured in units of newtons (N). In musical instruments, increasing the tension typically raises the pitch of the note produced by making the vibrations faster.

For instance, when designing a string instrument, the string must be tight enough to vibrate at a certain fundamental frequency. In this example, string S1 is tuned to produce the middle C note which has a frequency of 262 Hz. By using the formula \(f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}\), we can calculate the tension needed for this frequency when we know the vibrating string length and its linear density. The resultant high tension ensures the string vibrates at the required frequency, creating the desired pitch.

Linear Density of String

Linear density, symbolized by \(\mu\), is the measure of mass per unit length of a string and is expressed in units like grams per meter (g/m). It plays a critical role in determining the frequency of the sound wave produced by a string. The linear density along with tension and the length of the string dictates the rate at which the string will vibrate.

To find the mass of string S2 for our two-string instrument, we need to first calculate its linear density so that it will produce an A-sharp when vibrating. Once we've found the value \(\mu\), we can find the mass by multiplying the linear density by the length of the string. It's a step involving straightforward calculation, nonetheless vital for ensuring the string produces the intended note.

Frequency of Sound Wave

The frequency of a sound wave is directly related to the pitch of the sound we hear. It is measured in Hertz (Hz) and represents the number of vibrations per second. For a string instrument, when the string vibrates in its fundamental mode, the frequency is at its lowest value for that particular length and tension.

The known frequencies of musical notes allow us to design instruments with precision. Middle C has a frequency of 262 Hz, while A-sharp has a frequency of 466 Hz. Using the formula mentioned before, we rearrange it based on what we need to find, whether it is the tension for a known frequency or a frequency that will be produced by adjusting the tension or the length of the string. Subsequently, each variation requires a new calculation to ensure the correct note is produced.

Vibrating String Length

The length of a vibrating string determines the wavelengths of the sound waves it produces, which in turn affects the frequency and pitch of the sound. By adjusting the length of the string using a fret, as in the given exercise, we can alter the pitch it produces without changing the tension.

For example, when the fret is pressed against string S1 to produce a C-sharp note at a frequency of 277 Hz, this changes the effective vibrating length. Here we have calculated this new length (\(L'\)) using the same equation for fundamental frequency, but we rearranged it to solve for the new length based on the desired frequency. By controlling the vibrating string length, we enable the instrument to produce a variety of notes and extend its playable range.