A tube closed at one end has a vibrating diaphragm at the other end , which may be assumed to be a displacement node . It is found that when the frequency of the diaphragm is `2000 Hz` , a stationary wave pattern is set up in which the distance between adjacent nodes is `8 cm`. When the frequency is gradually reduced , the stationary wave pattern reappears at a frequency of `1600 Hz`. Calculate
i. the speed of sound in air ,
ii. the distance between adjacent nodes at a frequency of `1600 Hz`,
iii. the distance between the diaphragm and the closed end ,
iv. the next lower frequencies at which stationary wave patterns will be obtained.

Explanation

Since the node - to node distance is `lambda//2`
`lambda//2 = 0.08 or lambda = 0.16 m`
i. `c = n lambda`
`:. C = 2000 xx 0.16 = 320 m//s`
ii. `320 = 1600 xx lambda or lambda = 0.2 m`
`:.` Distance between nodes `= 0.2//2 = 0.1 m = 10 cm`
iii. Since there are nodes at the ends , the distance between the closed end the membranes must be exact integrals of `lambda//2`.
`:. I = n xx 0.16//2 and I = n' xx 0.2//2`
`rArr (n)/(n') = (5)/(4)`
When ` n = 5 , n' = 4`
`I = 5 xx 0.16//2 = 0.4 m = 40 cm`
iv. For the next lower frequency ` n = 3`
`:. 0.4 = 3 lambda//2 or , n = (320)/(0.8//3) = 1200 Hz`