91Ó°ÊÓ

Determination of the lengths wave could be identified based on the standing wave frequency.

At first, the need is to know the open-close tube so the formula of the frequency will be $fm=m\frac{v}{4L^{;}}$

Here m= 1,3, 5.. and the need are to find m.

based on the formula the frequency of the standing waves m could be expressed as,

$f=m\frac{v}{4L}⇒m=\frac{4f}{v}L$

The direct consequences are $$\begin{gathered} m_{min}=\frac{4f}{v}{L}_{min}=\frac{4f}{v}·0=\underline{0} \\ {m}_{max}=\frac{4f}{v}{L}_{max}=\frac{4.580}{343}·1=\underline{6.76} \end{gathered}$$

the od integers are 1,3,5 and so the L could be expressed asrole="math" localid="1649077298625" $f=m\frac{v}{4L}⇒\underline{L-m\frac{v}{4f}}$

The numerical are exists that $$\begin{gathered} L_1=\frac{343}{4·580}=\underline{0.148m,} \\ {L}_2=3\frac{343}{4·580}=\underline{0.444m,} \\ {L}_1=5\frac{343}{4·580}=\underline{0.739m.} \end{gathered}$$

These are the heights of the air that is present in the tube.

Here, the height of the water is similar to the tube length subtracted by the necessary air length that is calculated as L. Thus, the heights could be presented as descending order

$$\begin{gathered} h_1=1-L_3=1-0.739=\boxed{0.261m} \\ {h}_2=1-L_2=1-0.444=\boxed{0.556m} \\ {h}_1=1-L_1=1-0.148=\boxed{0.852m} \end{gathered}$$