Question

In a resonance tube experiment, a 1 m long pipe is resonated with a tuning fork of frequency 340 Hz. Initially the pipe is completely filled with water. A small hole is drilled very close to the bottom and water is allowed to leak. If the radii of the pipe and hole are 2 cm and Imm respectively, then the time interval between the occurrence of first two resonances is seconds. Neglect end correction.

Answer 1

we know, speed of sound = 340 m/s in air.

Let l is the length of air column corresponding to fundamental frequency then,

so, v/4l = 340 Hz

or, l = 340/4(340) = 0.25m

In closed pipe only odd harmonics are obtained. now $l_1,l_2,l_3....$ be the length of air column corresponding to 3rd, 5th , 7th.. harmonic respectively.

so, $l_1=3l=0.75m$ < 1m

$l_2=5l=1.25m$ > 1m

so, heights of water levels are (1-0.25) , (1 - 0.75).

or, heights of water levels are 0.75 m 0.25m

Let A and a be the area of cross-section of the pipe and hole respectively. Then

A = π(2 × 10^-2)² = 1.26 × 10^-3 m²

and a = π(10^-3)² = 3.14 × 10^-6 m²

now, from continuity,

$v_1A_1=v_2A_2$

velocity of efflux ,v = √{2gH}

or, $a\sqrt{2gH}=A\left(-\frac{dH}{dt}\right)$

or, $-\frac{dH}{dt}=1.11\times10^{-2}\sqrt{H}$

between two resonance water levels falls from 0.75m to 0.25m

or, $\int\limits^{0.25}_{0.75}\frac{dH}{\sqrt{H}}=-1.11\times10^{-2}\int\limits^t_0{dt}$

or, 2[√0.25 - √0.75] = -1.11 × 10^-2 t

or, 0.73 = 1.11 × 10^-2 t

t = 73/1.11= 65 sec

Answer 2

Since the tuning fork is in resonation with air area in the pipe shut down toward one side, the repeat is \[n=\frac{(2N-1)v}{4l}\] where N = 1, 2, 3 .... identifies with different technique for vibration putting n = 340Hz, v = 340 m/s, the length of air segment in the pipe can be \[l=\frac{(2N-1)340}{4\times 340}=\frac{(2N-1)}{4}m=\frac{(2N-1)\times 100}{4}cm\] For N = 1, 2, 3, ... we get l = 25 cm, 75 cm, 125 cm ... As the chamber is only 120 cm long, length of air section after water is poured in it may be 25 cm or 75 cm only, 125 cm is past the domain of creative ability, the relating length of water portion in the barrel will be (120 ? 25) cm = 95 cm or (120 ? 75) cm = 45 cm. In this manner least length of water area is 45 cm.