Question

SOLUTION (A) Find the frequencies if the pipe is open at both ends. _V 343 m/s Substitute into whole harmonics equation, with n = 1. 11-222(2.46 m) = 69.7 Hz Multiply to find the second and third harmonics. 12 - 27 - 139 Hz 13 = 3f7 - 209 Hz (B) How many harmonics lle between 20 Hz and 20000 Hz for this pipe? 343 m/s Set the frequency in the harmonics equation equal to 2.00 x 104 Hz and solve for n. 2. (2.46 m 2.00 x 104 Hz = 286 This works out to - 286.88, which must be truncated down (n = 287 gives a frequency over 2.00 x 10 Hz). (C) Find the frequencies for the pipe closed at one end. Apply the one end open harmonics equation with 1. 343 m/s 14L4(2.46 m) - 34.9 Hz The next two harmonics are odd multiples of the first: 13 - 31. - 105 HZ s = 51, - 175 Hz

Answer 1

Practice it (a) For pipe open at both end fi = Ž = 342 = 68.9542 f = 28, = 2x68.95 = 137.9 H2 fz = 35,= 3x68.95 = 206.85 HZ (b) fn =nf = 20000 ang at n = 290 ) (c) for pipe closed at one end F = - 342 4x2.48 = 34.48 Hz = 35,= 3x34.48 = 103.44 Hz fr = 56, = 5x34.48 = 172.4 Hz Exercise

Alt Gr Exercise ca) f. = 20fi 7 - L= V = 342 2x 362 = 0.472 m fovendone = 2f,= 2x362 = 724H2 (6) = 181.14 HZ Indone = V = 342 4x0.472 Sovertone = 3f, = 3x181.14 = 543.42 HZ fn = (2n-1) fi 20000 =(2n-1) 181.14 (2n-17 = 110.4 n=55