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
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.

Answer 1

The frequency
`f` = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Step-by-step explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank =
`(d)/(2)` =
`(44)/(2)` = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot =
`(d)/(2)` =
`(3.0)/(2)` = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder =
`(d)/(2)` =
`(2.0)/(2)` = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

`P_1+(1)/(2)pv_1^2+pgy_1=P_2+(1)/(2)pv^2_2+pgy_2`

`pgy_1=(1)/(2)pv^2_2 +pgy_2`

`v_2=√(2g(y_1-y_2))`

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ =
`√(2*9.8*(0.35-0))`

v₂ =
`\sqrt {6.86}`

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

`v_3 = (v_2r_2^2)/(v_3^2)`

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ =
`(2.62 m/s)*`
`((1.5mm^2)/(1.0mm^2) )`

v₃ = 0.0589 m/s

∴ velocity of the fluid in the cylinder = 0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

`f=(v_s)/(4(h-v_3t))`

where;

`v_s` = velocity of sound

h = height of the fluid

v₃ = velocity of the fluid in the cylinder

`f=(343)/(4(0.40-(0.0589)(0.4))`

`f= (343)/(0.6576)`

`f` = 521.59 Hz

∴ The frequency
`f` = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

`(df)/(dt)= (d)/(dt)((v_s)/(4)(h-v_3t)^(-1))`

`(df)/(dt)= -(v_s)/(4)(h-v_3t)^2(-v_3)`

`(df)/(dt)= (v_sv_3)/(4(h-v_3t)^2)`

where;

`v_s` (velocity of sound) = 343 m/s

v₃ (velocity of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

`(df)/(dt)= (343*0.0589)/(4(0.4-(0.0589*4.0))^2)`

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s when the cylinder has been filling for 4.0 s.