A `3.6m` long vertical pipe resonates with a source of frequency `212.5 Hz` when water level is at certain height in the pipe. Find the height of water level (from the bottom of the pipe) at which resonance occurs. Neglect end correction. Now , the pipe is filled to a height `H (~~ 3.6m)`. A small hole is drilled very close to its bottom and water is allowed to leak. Obtain an expression for the rate of fall of water level in the pipe as a function of `H`. If the radii of the pipe and the hole are `2 xx 10^(-2)m` and `1 xx 10^(-3)m` respectively, Calculate the time interval between the occurance of first two resonances. Speed of sound in air `340m//s` and `g = 10m//s^(2)`.

Answer

`h=3.2,2.4,1.6,0.8,v=5xx10^(-3)sqrt(5H),Deltat=80(4-2 sqrt(3))`

Explanation

Speed of sound
Let `l_(0)` be the length of air column corresponding to the fundamental frequency.
`(V)/(4l_(0))=212.5`
`rArr l_(0)=(V)/(4(212.5))=(340)/(4(212.5))=0.4m`

In cloed pipe only odd harmonics are obtained.
Now let `l_(1),l_(2),l_(3),l_(4)`. etc., be the lengths corresponding to the 3rd harmonic. 4th harmonic 7th harmonic etc. Then,
`3((V)/(4l_(1)))=212.5 rArr l_(1)=1.2 m`
`5((v)/(4l_(2)))=212.5 rArr l_(2)=2.0m`
and `7((V)/(4l_(3)))=212.5 rArr l_(3)=2.8`
`9((V)/(3l_(4)))=212.5 rArr l_(4)=3.6 m`
or heights of water level are
`(3.6-0.4)m,(3.6-1.3)m`,
`(3.6-2.0)m,(3.6-2.8)m`
`:.` Heights of water level are
`3.2,2.4m.1.6m and 0.8m`
Let A and a be the area of cross-section of the pipe and hole respectively. Then

`A=pi(2xx10^(-2))^(2)=1.26xx10^(-3)m^(2)`
and `a_(1)=pi(10^(-3))%(2)=3.14xx10^(-6)m^(2)`
Velocity of efflux
Continuity equation at 1 and 2 gives :
`asqrt(2gH)=A((-dH)/(dt))`
`:.` Rate of fall of water level in the pipe
`(-(dH)/(dt))=(a)/(A)sqrt(2gH)`
Substituting the values we get
`(-dH)/(dt)=(3.14xx10^(-6))/(1.26xx10^(-3))sqrt(2xx10xxH)`
`rArr -(dH)/(dt)=(1.11xx10^(-2))sqrt(H)`
Between first two resonances, the water level falls from 3.2 m to 2.4 m
`:.(dH)/(sqrt(H))=-(1.11xx10^(-2))dt`
`rArr underset(3.2)overset(2.4)(int) (dH)/(sqrt(H))=-(1.11 xx10^(-2))underset(0)overset(t)(int)dt`
`rArr 2[sqrt(2.4)-sqrt(3.2)]=-(1.11xx10^(-2))t`
`rArrt ~~ 43s`.