For a certain organ pipe, the first three resonance frequencies are in the ratio of 1:3:5 respectively. If the frequency of fifth harmonic is 405 Hz and the speed of sound in air is 324 ms-1 the length of the organ pipe is ___ m.
To solve this problem, we need to find the length of the organ pipe when given that its resonance frequencies are in the ratio 1:3:5 and the frequency of the fifth harmonic is 405 Hz. We know the first three resonant frequencies of an organ pipe correspond to the harmonics, and for a closed organ pipe these are usually the odd harmonics. If the resonant frequencies are in the ratio 1:3:5, then this indicates that the harmonics are the first, third, and fifth harmonics. Given the fifth harmonic frequency \(f_5\) is 405 Hz, and the speed of sound \(v\) is 324 m/s, we can use the formula for the frequency of the \(n\)th harmonic in a closed pipe: \(f_n=\frac{nv}{4L}\), where \(n\) is the harmonic number and \(L\) is the length of the pipe. For the fifth harmonic: \[f_5=\frac{5\cdot324}{4L}=405\]. Solve for \(L\): \[405=\frac{1620}{4L}\] \[\Rightarrow L=\frac{1620}{4\cdot405}\] \[\Rightarrow L=\frac{1620}{1620}=1\text{ m}\]. Thus, the length of the organ pipe is 1 m, which is within the range 1,1 as expected.
