The speed of sound in air at room temperature is 350 m/s. A pipe is 35

1.	The speed of sound in air at room temperature is 350 m/s. A pipe is 35 cm in length. Find the frequency of the third overtone in the pipe when it is (i) closed at one end (ii) open at both ends. Ignore the end correction.
Answer» Data : v = 350 m/s, L = 35 cm = 35 × 10^-2 m (i) For a pipe closed at one end, the fundamental frequency is n_c = \(\frac v{4L}\) = \(\frac{350}{4\times35\times10^{-2}}\) = 250 Hz As only odd harmonics are present in this case, the frequency of pth overtone is n_p = (p × 2 + 1) n_c ∴ The frequency of the 3rd overtone is n₃ = (3 × 2 + 1)n_c = 7n_c = 7 × 250 = 1750 Hz (ii) For a pipe open at both ends, the fundamental frequency is n_O = \(\frac v{2L}\) = \(\frac{350}{2\times35\times10^{-2}}\) = 500 Hz In this case, all harmonics are present. ∴ The frequency of the pth overtone is n_p= (p + 1) n_o ∴ The frequency of the 3rd overtone is n₃ = (3 + 1) n_o = 4n_o = 4 × 500 = 2000 Hz

The speed of sound in air at room temperature is 350 m/s. A pipe is 35 cm in length. Find the frequency of the third overtone in the pipe when it is (i) closed at one end (ii) open at both ends. Ignore the end correction.

Answer»

Data : v = 350 m/s, L = 35 cm = 35 × 10^-2 m

(i) For a pipe closed at one end, the fundamental frequency is

n_c = \(\frac v{4L}\) = \(\frac{350}{4\times35\times10^{-2}}\) = 250 Hz

As only odd harmonics are present in this case, the frequency of pth overtone is n_p = (p × 2 + 1) n_c

∴ The frequency of the 3rd overtone is

n₃ = (3 × 2 + 1)n_c = 7n_c = 7 × 250 = 1750 Hz

(ii) For a pipe open at both ends, the fundamental frequency is

n_O = \(\frac v{2L}\) = \(\frac{350}{2\times35\times10^{-2}}\) = 500 Hz

In this case, all harmonics are present.

∴ The frequency of the pth overtone is

n_p= (p + 1) n_o

∴ The frequency of the 3rd overtone is n₃ = (3 + 1) n_o = 4n_o = 4 × 500 = 2000 Hz