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Q. 56

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Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Found in: Page 486

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Short Answer

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.

a) The frequency is 523 Hz

b) The frequency is 8350 Hz/s

See the step by step solution

Step by Step Solution

Step 1: The concept of the fundamental frequency 

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

Step 2: Identification of the frequency 

The frequency and 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

In this context, it will be

Then the description of the function will be

The total length of the tube is

The volumetric flow rate of water will be Q --so the volume flowing per unit time is

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

is the diameter of the tube so the equation is

Step 3: 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

The h is the height of the tank so the subtraction would be

The remaining height as the function of time as

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

The formula is by the subtraction will be

The provide numerical scenario is

The subtraction of the diameters could be conducted as the units cancels out which is and it is 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

Thus, the outcome is

The given time is

In this case, the height is approximately and responsible for 65% of filling.

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