Presentation is loading. Please wait.

Presentation is loading. Please wait.

v = f fn = nf1  = 5558 m 343 = 5558(f) f = Hz 325th harmonic

Similar presentations


Presentation on theme: "v = f fn = nf1  = 5558 m 343 = 5558(f) f = Hz 325th harmonic"— Presentation transcript:

1 v = f fn = nf1  = 5558 m 343 = 5558(f) f = 0.0617 Hz 325th harmonic
Given: L = 2779 m n = 1 (fundamental frequency) v = 343 m/s 1) (open air column) a) v = f  = 5558 m 343 = 5558(f) f = Hz b) fn = nf1 325th harmonic 20 = n(0.0167)

2 v = f v = f  = 0.96 m 343 = 0.96(f) f = 357.3 Hz = 476.4 Hz
Given: L = 48 cm = 0.48 m n = 1 (fundamental frequency) v = 343 m/s 2) (open air column) v = f  = 0.96 m 343 = 0.96(f) f = Hz = Hz L = 0.36 m v = f 12 cm from the end 343 = (476.4)  = 0.72 m

3 v = f open pipe f2 = 306.2 Hz; f3 = 459.3 Hz; f4 = 612.4 Hz
3) Given: L = 112 cm = 1.12m n = 1 (fundamental frequency) v = 343 m/s open pipe f2 = Hz; f3 = Hz; f4 = Hz (open air column) closed pipe f3 = Hz; f5 = Hz; f7 = Hz open = 2.24 m (closed air column) closed = 4.48 m v = f open pipe f1 = Hz 343 = 2.24(f) closed pipe f1 = 76.6 Hz 343 = 4.48(f) fn = nf1 open pipe next three harmonics = 2, 3, 4 closed pipe next three harmonics = 3, 5, 7

4 60 Hz 100 Hz 140 Hz. f3 f5 f7 20 Hz These differ by how much ? 40 Hz
4) (closed air column) These differ by how much ? 40 Hz 60 Hz Hz Hz. f3 f5 f7 These differ by how much ? 40 Hz By how much less would you expect the harmonic before 60 Hz to be? 40 Hz What is the frequency of the harmonic before 60 Hz? I know this is the fundamental frequency since 40 Hz below this would be negative 20 Hz

5 closed air column 88 Hz 264 Hz 440 Hz 616 Hz. f3 f5 f7 88 Hz
5) 88 Hz These differ by how much ? 176 Hz 264 Hz Hz Hz. f3 f5 f7 These differ by how much ? 176 Hz By how much less would you expect the harmonic before 264 Hz to be? 176 Hz What is the frequency of the harmonic before 264 Hz? I know this is the fundamental frequency since 176 Hz below this would be negative 88 Hz

6 v = f v = f 343 = (262)  = 1.309 m 337 = 1.309(f) L = 65.5 cm
Given: f = 262 Hz n = 1 (fundamental frequency) v = 343 m/s 6) (open air column) v = f 343 = (262)  = m v = f 337 = 1.309(f) L = 65.5 cm f = Hz

7 v = f 343 = (294)  = 1.16 m L = 0.58 m 7.5 cm from the end Given:
f = 294 Hz n = 1 (fundamental frequency) v = 343 m/s 7) (open air column) v = f 343 = (294)  = 1.16 m L = 0.58 m 7.5 cm from the end


Download ppt "v = f fn = nf1  = 5558 m 343 = 5558(f) f = Hz 325th harmonic"

Similar presentations


Ads by Google