One of the harmonic frequendies of tube A with two open ends is 1404 Hz. The nex
ID: 1872957 • Letter: O
Question
One of the harmonic frequendies of tube A with two open ends is 1404 Hz. The next-highest harmonic frequerncy is 1482 H2. (a) What harmonic frequency is next highest after the harmonic frequency 458 Hz? (b) What is the number of this next-highest harmonic? One of the harmonic frequencies of tube 8 with orly one open end is 1560 Hz. The next-highest harmonit frequency is 1900 Hz. (c) What harmit frequency is riext highest after the harmonic frequency 940 Hz? (d) What is the number of this next-highest haonic? (a) Numbor (b) Numbor Number (d) Number unitsExplanation / Answer
The frequency of the open pipe is fm = 1404 Hz
The ferquency of the next harmonic is fm+1 = 1482 Hz
For an open pipe the difference in the frequency of two consecutive resonating coloumns is given by
fm+1 - fm = v/2L
v is the speed of sound
L is the length of the coloumn
Therefore
v/2L = 1482 Hz - 1404 Hz
v/2L = 78 Hz
a)
The frequency of the next resonating coloumn for 468 HZ is
f' = 468 Hz + v/2L
f' = 468 Hz + 78 Hz
f' = 546 Hz
b)
The number of the harmonic is m = 546 Hz/78 Hz
= 7
The frequency of the closed pipe is fm = 1560 Hz
The ferquency of the next harmonic is fm+1 = 1800 Hz
For an open pipe the difference in the frequency of two consecutive resonating coloumns is given by
fm+1 - fm = v/2L
v is the speed of sound
L is the length of the coloumn
Therefore
v/2L = 1800 Hz - 1560 Hz
v/2L = 240 Hz
The frequency of the next resonating coloumn for 840 Hz is
f' = 840 Hz + v/2L
f' = 840 Hz + 240 Hz
f' = 1080 Hz
d)
The number of the harmonic is
m = 840 Hz /120 Hz
m = 7
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