The number of possible natural oscillations of air column in a pipe closed at one end of length 85 ~cm whose frequencies lie below 1250 ~Hz are (velocity of sound =340 ~ms ⁻¹ )

Step-by-step solution

For a pipe closed at one end, the frequencies of natural oscillations are given by the formula f_n = (2n-1)v 4L , where n = 1, 2, 3, is the harmonic number, v is the velocity of sound, and L is the length of the pipe. Given v = 340 m/s and L = 85 cm = 0.85 m . Substituting the values into the formula: f_n = (2n-1) 340 4 0.85 = (2n-1) 340 3.4 = (2n-1) 100 Hz . We require f_n (2n-1) 100 2n-1 2n n Since n must be a natural number, the possible values for n are 1, 2, 3, 4, 5, 6 . The number of possible natural oscillat