In a closed organ pipe, the frequency of the fundamental note is 30 Hz. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz. If the organ pipe has a cross-sectional area of 2 cm\(^2\), the amount of water poured in the organ tube is \( \_\_\_\_\_\_ \) g. (Take speed of sound in air as 330 m/s)
Correct Answer: 400
Approach Solution - 1
The problem asks for the mass of water poured into a closed organ pipe, which causes its fundamental frequency to change from 30 Hz to 110 Hz. We are given the cross-sectional area of the pipe and the speed of sound in air.
Concept Used:
The solution is based on the formula for the fundamental frequency of a closed organ pipe (a pipe closed at one end and open at the other).
The fundamental frequency (\(f\)) is given by:
\[ f = \frac{v}{4L} \]where \(v\) is the speed of sound in air, and \(L\) is the length of the air column inside the pipe.
When water is poured into the pipe, it effectively shortens the length of the resonating air column, thereby increasing its fundamental frequency. The volume of the water added can be calculated from the change in the length of the air column and the cross-sectional area. The mass can then be found using the density of water (\(\rho_{water} = 1 \, \text{g/cm}^3\)).
\[ \text{Volume} = \text{Area} \times \text{Height} \] \[ \text{Mass} = \text{Volume} \times \text{Density} \]Step-by-Step Solution:
Step 1: Calculate the initial length of the air column (\(L_1\)).
The initial fundamental frequency is \(f_1 = 30 \, \text{Hz}\), and the speed of sound is \(v = 330 \, \text{m/s}\). Using the formula for the fundamental frequency:
\[ L_1 = \frac{v}{4f_1} \] \[ L_1 = \frac{330 \, \text{m/s}}{4 \times 30 \, \text{Hz}} = \frac{330}{120} \, \text{m} = 2.75 \, \text{m} \]This is the initial length of the organ pipe.
Step 2: Calculate the new length of the air column (\(L_2\)) after adding water.
The new fundamental frequency is \(f_2 = 110 \, \text{Hz}\).
\[ L_2 = \frac{v}{4f_2} \] \[ L_2 = \frac{330 \, \text{m/s}}{4 \times 110 \, \text{Hz}} = \frac{330}{440} \, \text{m} = \frac{3}{4} \, \text{m} = 0.75 \, \text{m} \]This is the length of the air column remaining after the water has been poured in.
Step 3: Determine the height of the water column in the pipe.
The reduction in the length of the air column is due to the water that has been poured in. Therefore, the height of the water column (\(h_{water}\)) is the difference between the initial and final lengths of the air column.
\[ h_{water} = L_1 - L_2 = 2.75 \, \text{m} - 0.75 \, \text{m} = 2.0 \, \text{m} \]Step 4: Calculate the volume of the water poured into the pipe.
The cross-sectional area is given as \(A = 2 \, \text{cm}^2\). It is convenient to convert the height of the water to centimeters.
\[ h_{water} = 2.0 \, \text{m} = 200 \, \text{cm} \]The volume of the water is:
\[ V_{water} = A \times h_{water} = (2 \, \text{cm}^2) \times (200 \, \text{cm}) = 400 \, \text{cm}^3 \]Final Computation & Result:
Step 5: Calculate the mass of the water.
The density of water is \(\rho_{water} = 1 \, \text{g/cm}^3\).
\[ \text{Mass} = \text{Volume} \times \text{Density} \] \[ m_{water} = 400 \, \text{cm}^3 \times 1 \, \text{g/cm}^3 = 400 \, \text{g} \]The amount of water poured into the organ tube is 400 g.
Approach Solution -2
Step 1: Calculate the Initial Length of Air Column:
For a closed organ pipe, the fundamental frequency is given by:
f = V / 4ℓ1
where f = 30 Hz and V = 330 m/s.
Solving for ℓ1:
ℓ1 = V / 4 × f = 330 / (4 × 30) = 330 / 120 = 11 / 4 m = 2.75 m
Step 2: Calculate the New Length of Air Column:
When the frequency increases to 110 Hz:
f' = V / 4ℓ2
Solving for ℓ2:
ℓ2 = V / 4 × f' = 330 / (4 × 110) = 330 / 440 = 3 / 4 m = 0.75 m
Step 3: Determine the Change in Length:
Δℓ = ℓ1 - ℓ2 = 2.75 - 0.75 = 2 m
Step 4: Calculate the Volume of Water Added:
The volume of water added corresponds to the volume of the air column displaced:
Change in volume = A × Δℓ = 2 cm² × 200 cm = 400 cm³
Step 5: Convert Volume to Mass:
Given that the density of water ρ = 1 g/cm³:
M = ρ × Volume = 1 × 400 = 400 g
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