A tuning fork produces a signal of 530 Hz. What frequency is...

Question

1) A tuning fork produces a signal of 530 Hz. What frequency is

(a) two octaves below this frequency?

(b) three octaves above this frequency?

2) Certain pipe organs can act like air columns that are open at both ends. If the speed of sound is 343 m/s, how long must a pipe be to produce a frequency of 25 Hz (the lower hearing range of most adults)?

3) The "standard" note (A4 = 440 Hz) on a harp has a string length of 75 cm. How long would the same string have to be to produce middle C (261.6 Hz)?

4) An air column closed at on end, with a length of 25.3 cm, resonates in the first overtone. If the frequency of the sound wave is 1024 Hz, find the air temperature inside the column.

Please show full solutions. Thank you for your help!

Answer 1

1)

(a) Two octaves below 530 Hz is 132.5 Hz. To find this, we can use the formula:

f2 = f1 / 2^n

where f1 is the original frequency, n is the number of octaves, and f2 is the new frequency. In this case, we want to find f2 when n = 2:

f2 = 530 / 2^2 = 132.5 Hz

(b) Three octaves above 530 Hz is 4240 Hz. Using the same formula as above:

f2 = 530 * 2^3 = 4240 Hz

2)

The formula for the fundamental frequency of an open pipe is:

f = nv/2L

where n is the harmonic number (1 for the fundamental), v is the speed of sound, and L is the length of the pipe. Solving for L:

L = nv/2f

Substituting the given values:

L = (1)(343 m/s) / (2*25 Hz) = 6.86 m

Therefore, the pipe must be 6.86 meters long to produce a frequency of 25 Hz.

3)

The formula for the fundamental frequency of a vibrating string is:

f = nv/2L

where n is the harmonic number (1 for the fundamental), v is the speed of the wave, and L is the length of the string. Solving for L:

L = nv/2f

Substituting the given values:

L = (1)(v) / (2f)

To find the length of the string that produces middle C (261.6 Hz), we can use the standard note as a reference point:

L2 = (440 m/s) / (2 * 261.6 Hz) = 0.84 m

Therefore, the length of the string that produces middle C is 0.84 meters.

4)

The formula for the resonant frequencies of an air column closed at one end is:

f = (2n - 1)v/4L

where n is the harmonic number, v is the speed of sound, and L is the length of the column. For the first overtone, n = 2. Solving for L:

L = (2n - 1)v/4f

Substituting the given values:

L = (2*2 - 1)(343 m/s) / (4*1024 Hz) = 0.253 m

Therefore, the length of the air column is 0.253 meters. To find the air temperature, we can use the ideal gas law:

PV = nRT

where P is the pressure inside the column, V is the volume, n is the number of moles of air, R is the gas constant, and T is the temperature. Since the column is closed at one end, the pressure at the closed end is high and the pressure at the open end is atmospheric pressure. Therefore, we can assume that the pressure inside the column is roughly twice the atmospheric pressure (since the column is in the first overtone). Substituting the given values:

(2 * 1.013 * 10^5 Pa) * (pi * (0.0253 m)^2 / 4) = n * 8.31 J/mol*K * T

Solving for T:

T = [(2 * 1.013 * 10^5 Pa) * (pi * (0.0253 m)^2 / 4)] / (n * 8.31 J/mol*K)

Assuming that the air inside the column behaves like an ideal gas, we can use the molar mass of air to find the number of moles of air:

n = (m / M)

where m is the mass of the air and M is the molar mass of air. Assuming the air has a density of 1.2 kg/m^3 and a volume of 0.253 m^3 (the volume of the air column), we can find the mass of the air:

m = density * volume = (1.2 kg/m^3) * (0.253 m^3) = 0.304 kg

The molar mass of air is approximately 28.97 g/mol. Converting the mass of air to moles:

n = (0.304 kg) / (28.97 g/mol) = 0.0105 mol

Substituting the values for n and solving for T:

T = [(2 * 1.013 * 10^5 Pa) * (pi * (0.0253 m)^2 / 4)] / (0.0105 mol * 8.31 J/mol*K) = 291 K

Therefore, the air temperature inside the column is approximately 291 K (or 18.9°C).