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ID: 1493105 • Letter: 9
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94% C4 Sun Apr 10 10 54 44 PM a E Chrome File Edit View History Bookmarks People Window Help Ogoogle maps Google C My Internships IChood C Physics question I Ch A Josh Chapter 17 Linear S C Books ke- c www.webassign.net 13047796 /web/Student/Assignment-Responses/last?dep Apple YouTube home Watch L Sports E O tech EN9 faculty pnc.edu/pte Bookmarks Destiny LFG Net I App Screen Shot PM 2016-04...0.08 PM 9. -12 points CJ10 17.P.044 My Notes Ask Your Teacher The range of human hearing is roughly from twenty hertz to twenty kilohertz. Based on these limits and a value of 320 m/s for the speed of sound, what are the lengths of the longest and shortest pipes (open at both ends and producing sound at their fundamental Screen Shot frequencies) that you expect to find in a pipe organ? PM 2016-04...0.20 PM shortest pipe longest pipe Additional Materials o your Screen Shot Section 17.7 2 PM 2016-04...0.26 PM you a Screen Shot PM 2016-04...0.30 PM 10. -l2 points CJ10 17.P.047 My Notes Ask Your Teacher A tube is open only at one end. A certain harmonic produced by the tube has a frequency o 425 Hz. The next higher harmonic has a his equency of 546 Hz. The speed of sound in a s 343 m/s s the eger n that describes the armonic whose a) Wha requency is 425 Hz? Screen Shot PM 2016-04...0.40 PM (b) What is the length of the tube? Additional Materials Screen Shot PM 2016-03...4.06 PM Section 17.7Explanation / Answer
9) let L is the length of the pipe.
fundamental frequency of open pipe,
f = v/(2*L)
f_max = v/(2*L_shortest)
==> L_shortest = v/(2*f_max)
= 320/(2*20000)
= 0.008 m or 8 mm
f_min = v/(2*L_longest)
==> L_longest = v/(2*f_min)
= 320/(2*20)
= 8 m
10)
in a closed pipe,
fn - fn-1 = 2*f1
f1 = (fn - fn-1)/2
= (546 - 425)/2
= 60.5 hz
a) fn = n*f1
==> n = 425/60.5
= 7
f1 = v/(4*L)
==> L = v/(4*f1)
= 343/(4*60.5)
= 1.417 m
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