91Ó°ÊÓ

• Length of glass tube ,$\mathrm{L}=1.\text{00m}$
• Frequency of the tuning fork ,$\mathrm{f}=68\text{6Hz}$

Apply the formula for the harmonics (frequency) for a closed organ pipe to find the different positions of water levels.

Formula is as follow:

$\mathrm{f}=\left(\mathrm{n}−\frac{1}{2}\right)\left(\frac{\mathrm{v}}{2\mathrm{X}}\right)$

Here, $\mathrm{f}$is frequency,$\mathrm{x}$ is the position and$\mathrm{v}$is the velocity.

Frequency (resonance) of closed pipe is defined as,

$\mathrm{f}=\left(\mathrm{n}−\frac{1}{2}\right)\left(\frac{\mathrm{v}}{2\mathrm{X}}\right)$

Here, $\mathrm{x}$is the distance from the top of the tube,

$68\text{6Hz}=\left(\mathrm{n}−\frac{1}{2}\right)\left(\frac{343\text{ }\mathrm{m}/\mathrm{s}}{2\mathrm{X}}\right)$

$\mathrm{X}=\left(\mathrm{n}−\frac{1}{2}\right)\left(\frac{1}{4}\right)$

When,$\mathrm{n}=1$,

$\begin{array}{c}\mathrm{X}=\left(1−\frac{1}{2}\right)\left(\frac{1}{4}\right)\\ =0.12\text{5m}\end{array}$

When, $\mathrm{n}=2$

$\begin{array}{c}\mathrm{X}=\left(2−\frac{1}{2}\right)\left(\frac{1}{4}\right)\\ =0.375\text{ }\mathrm{m}\end{array}$

When,$\mathrm{n}=3$

$\begin{array}{c}\mathrm{X}=\left(3−\frac{1}{2}\right)\left(\frac{1}{4}\right)\\ =0.62\text{5m}\end{array}$

When,$\mathrm{n}=4$

$\begin{array}{c}\mathrm{X}=\left(4−\frac{1}{2}\right)\left(\frac{1}{4}\right)\\ =0.875\text{″¾}\end{array}$

Hence, these are the different positions of water.

For the height of water, measure the height from the bottom of the tube,

$\begin{array}{c}{\mathrm{h}}_1=1\text{″¾}−0.875\text{m}\\ =0.\text{125m}\end{array}$

Hence, the least water heights in the tube for resonance to occur is $0.\text{125m}$.

Second least height

$\begin{array}{c}{\mathrm{h}}_2=1\text{″¾}−0.62\text{5m}\\ =0.37\text{5m}\end{array}$

Hence, the second least water heights in the tube for resonance to occur is $0.37\text{5m}$.