The air in a pipe resonates at 150 Hz and 750 Hz,one of these resonances being the fundamental.If the pipe is open at both ends,how many resonances are between the two given ones,and if the pipe is closed at one end,how many resonances are between the two given ones?
A)open: 3;closed: 1
B)open: 1;closed: 3
C)open: 2;closed: 0
D)open: 0;closed: 2

Question

Quizplus · Accepted Answer

For an open pipe, the resonant frequencies are given by f = n*v/2L, where n is an integer, v is the speed of sound, and L is the length of the pipe. The given frequencies correspond to n=1 and n=3. The number of resonances between them is therefore 3-1=2.
For a closed pipe, the resonant frequencies are given by f = n*v/4L, where n is an odd integer. The given frequencies correspond to n=3 and n=7. The number of resonances between them is therefore (7-3)/2=2/2=1.

The Air in a Pipe Resonates at 150 Hz and 750