Question

A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and water begins to flow out. The water falls into a 2.0-cm-diameter, 40-cm-tall glass cylinder. As the water falls and creates noise, resonance causes the column of air in the cylinder to produce a tone at the column’s fundamental frequency. What are (a) the frequency and (b) the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s? You can assume that the height of the water in the tank does not appreciably change in 4.0 s.

Short answer

a) The frequency is 523 Hz

b) The frequency is 8350 Hz/s

Step-by-step solution

The concept of the fundamental frequency 

The fundamental frequency is the lowest frequency in a periodic waveform.

Identification of the frequency 

The frequencyand the time are presented as f and t. The determination of the length is y which remains empty in time (t) and the tube length is L. Thus, the frequency is $f=\frac{v}{4L}$

In this context, it will be $4\left(t\right)=\frac{v}{4y\left(t\right)}$

Then the description of the function will be $f\left(t\right)$

The total length of the tube is $L=0.4$

The volumetric flow rate of water will be Q --so the volume flowing per unit time is $(L-y)A=QT$

Here, A is the cross-sectional area of the tube which has both sides \filled with the volume of the water.

The expression of the cross-sectional area is $A=\frac{\pi D^2}{4}$

$D=0.02m$is the diameter of the tube so the equation is$$\begin{gathered} L-y=\frac{4QT}{{\mathrm{\pi D}}^2} \\ y=L-\frac{4QT}{{\mathrm{\pi D}}^2} \end{gathered}$$

Assuming the height of the water

The volumetric flow rate is Q, the cross-section area is S and the speed is v.

Thus, Q=Sv

Based on the hydrodynamics it can be expressed as $pgh=\frac{p{v}^2}{2}\Rightarrow v=\sqrt{2gh},$

The h is the height of the tank so the subtraction would be $Q=S\sqrt{2gh}=\frac{\mathrm{\pi }{\varnothing }^2}{4}\sqrt{2gh}$

The remaining height as the function of time as $\frac{\underline{y\left(t\right)}=L-{\left({\displaystyle \frac{\varnothing }{D}}\right)}^{2}t\sqrt{2gh}}{}$

Because of the flow rate but the speed gets higher and the cross-section gets smaller.

The formula is by the subtraction will be $\boxed{f\left(t\right)=\frac{v}{4(L-{\left({\displaystyle \frac{\varnothing }{D}}\right)}^{2}t\sqrt{2gh})}}$

The provide numerical scenario is $f\left(4\right)=\frac{343}{4(0.4-{\left({\displaystyle \frac{3}{30}}\right)}^2·4\sqrt{2·9.8·0.35})}$

The subtraction of the diameters could be conducted as the units cancels out which is $f\left(4\right)=523Hz$and itis the ultimate result.

On the other hand, The frequency could be determined based on the derived frequency. therefore, the calculation of the frequency depends on the derived from the point of interest where $t=4s$

Thus, the outcome is $\frac{df}{dt}=\frac{v}{4}(-\frac{{\left({\displaystyle \frac{\varnothing }{D}}\right)}^2\sqrt{2gh}}{L-{\left(\frac{\varnothing }{D}\right)}^{2}t\sqrt{2gh}})$

The given time is $\boxed{\frac{df}{dt}\left(4\right)=8350Hz/s}$

In this case, the height is approximately$y\approx 16cm$and responsible for 65% of filling.