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.

Answer

10 cm

Explanation

At frequency 1600 hz wavelength is
`lambda = (v)/(n) = (320)/(1600)` = 0.2 m
distance between adjascent nodes is
`(lambda)/(2)` = 0.1m