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

1200 Hz, 800 Hz and 400 Hz

Explanation

All resonating frequency will be from fundamental to 2000Hz
`n_(R) =` 400 Hz, 800Hz, 1200 Hz, 1600 Hz & 2000 Hz