Question

An air-filled pipe is found to have successive harmonics at 980 Hz , 1260 Hz , and 1540 Hz . It is unknown whether harmonics below 980 Hz and above 1540 Hz exist in the pipe. What is the length of the pipe

Answer 1

The harmonics given indicate that the pipe's fundamental frequency is 700 Hz. By using the formula for the fundamental frequency of a pipe closed at one end, the length of the pipe is calculated to be 0.1225 meters or 12.25 cm.

Step-by-step explanation:

The given frequencies correspond to the harmonics of the air-filled pipe which is closed at one end. Because the pipe can only support odd harmonics, the frequencies given must be the fundamental frequency (1st harmonic) and the 3rd and 5th harmonics. To find the fundamental harmonic, we look at the difference between successive harmonics. The difference between the 3rd and 5th harmonics is 1540 Hz - 1260 Hz = 280 Hz; between the 1st and 3rd harmonics is 1260 Hz - 980 Hz = 280 Hz. Thus, the fundamental frequency is 980 Hz - 280 Hz = 700 Hz.

Using the formula for the fundamental frequency (f) of a pipe closed at one end, f = v/(4L), where v is the speed of sound (assuming 343 m/s at room temperature) and L is the length of the pipe, we can solve for L:

L = v/(4f) = 343 m/s / (4 * 700 Hz) = 0.1225 m

Therefore, the length of the air-filled pipe is 0.1225 meters or 12.25 cm.

Answer 2

L = 0.7 m

Step-by-step explanation:

This is a resonance exercise, in this case the air-filled pipe is open at both ends, therefore we have bellies at these points.

λ / 2 = L 1st harmonic

λ = L 2nd harmonic

λ = 2L / 3 3rd harmonic

λ = 2L / n n -th harmonic

the speed of sound is related to wavelength and frequencies

v =λ f

f = v /λ

we substitute

f = v n / 2L

the speed of sound in air is v = 343 m / s

suppose that the frequency of f = 980Hz occurs in harmonic n

f₁ = v n / 2L

f₂ = v (n + 1) / 2L

f₃ = v (n + 2) / 2L

we substitute the values

2 980/343 = n / L

2 1260/343 = (n + 1) / L

2 1540/343 = (n + 2) / L

we have three equations, let's use the first two

5.714 = n / L

7.347 = (n + 1) / L

we solve for L and match the expressions

n / 5,714 = (n + 1) / 7,347

7,347 n = 5,714 (n + 1)

n (7,347 -5,714) = 5,714

n = 5,714 / 1,633

n = 3.5

as the number n must be integers n = 4 we substitute in the first equation

L = n / 5,714

L = 0.7 m