Question

One string of a certain musical instrument is 80.0cm long and has a mass of 8.72 g. It is being played in a room where the speed of sound is 344m/s To what tensionmust you adjust the string so that, when vibrating in its second overtone, it produces sound of wavelength 0.768m ? (Assume that the breaking stress of the wire isvery large and isn’t exceeded.). What frequency sound does this string produce in its fundamental mode of vibration?

Answer 1

Concepts and reason

The concepts used to solve this problem are linear density, fundamental frequency, and overtone.

Initially, use the mass and length to calculate the linear mass density of the string.

Then, use the concept of standing wave for the second overtone to calculate the wavelength.

Then, use the velocity of sound and the wavelength of sound to calculate the frequency of the sound wave.

Then, use the frequency of the sound wave and the wavelength to calculate the speed of a wave in the string.

Then, use the velocity and the linear mass density to calculate the tension in the string.

Finally, use the concept of fundamental frequency to calculate the frequency of sound in the second overtone.

Fundamentals

The expression that relates velocity of the wave, tension on the string, and linear mass density is as follows:

Here, the tension on the string is and the linear mass density is .

The linear mass density is given by the following expression:

Here, is the mass and is the length of the string.

The formula to calculate the velocity of a wave in a string is as follows:

Here, the tension in the string is and the linear density of the string is .

The expression for the relation between the fundamental frequency and the harmonic frequency is as follows:

Here, the frequency of the harmonic is , the number of harmonic is , and the fundamental frequency is .

The linear mass density is given by the following expression:

Substitute for and for .

The expression for the wavelength of the standing wave for the second overtone is as follows:

Substitute for .

The expression for the frequency of the sound wave is as follows:

Here, frequency of the sound wave is , velocity of the sound wave is , and wavelength of the sound wave is .

Substitute for and for .

The expression for the speed of the wave in the string is as follows:

Substitute for and for .

The expression that relates the velocity of the wave, tension on the string, and linear mass density is as follows:

Rearrange the expression for .

Substitute for and for .

The expression for the relation between the fundamental frequency and the harmonic frequency is as follows:

For the first overtone:

Substitute for .

Rearrange the expression for .

Use for .

Substitute for .

Ans:

The tension in the string is .

The fundamental frequency is .

Answer 2

ing part B will be easier, so we will do that first.
The frequency of the second overtone will be
344 m/s / 3.35e-2 m = 1.03e4 Hz. The fundamental will be 1/3 this frequency, or 3.42e3 Hz, which turns out to be a note about midway between A flat and Anatural three octaves and a sixth above middle C.

The fundamental will have a wavelength twice as long as the string and the velocity will be the wavelength times the frequency. We can therefore find thevelocity of propagation along the string, which is given by the formula
v = v(T/µ)
where
v is the velocity of propagation,
T is the tension in the string, and
µ is the mass per unit length of the string.
The tension is therefore
T = v²µ
= v²m/l
where m is the mass of the string and l is the length of the string. But v = (2l)f, where f is the fundamental frequency, so
T = 4lmf²
= 4 * 0.800 m * 8.76e-3 kg * (344 m/s / (3*3.35e-2 m))² N
= 3.28e5 N

Answer 3

Frequency of the sound in air:
(nu = V_air / lambda ) = 447.92 Hz

Frequency of sound wave on string:
(nu ' = sqrt{TL/m} / 2L)
where, T = Tension in the string
m = mass of the string
L = length of the string
As we know that (nu' = nu) for equilibrium condition between the sound transmission from string to air by solving,
T = 5831.63 kg/(m-s^2)
The Fundamental (nu =) 447.92 Hz

Answer 4

The frequency of the second overtone is the velocity of propagation divided by the wavelength. Because the second overtone is the third harmonic, itvibrates at three times the frequency of the fundamental and the wavelength on the string will be exactly 2/3 the length of the string.

Answer 5

1399.59 N
b.223.958 Hz