A closed organ pipe has fundamental frequency of 1.5 kHz. The number of overtones that can be distinctly heard by a person with this organ pipe will be : (Assume that the highest frequency a person can hear is 20,000 Hz)

Explanation

If a closed pipe vibration in Nth mode then frequency of vibration `n=((2N-1)v)/(4l)=(2N-1)n_(1)`
(where `n_(1)=` fundamental frequency of vibration)
Hence 20,000`=(2N-1)xx1500`
`impliesN=7.1~~7`
`:.` Number of over tones =(No. of mode of vibration)-1
`=7-1=6`