Question

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)

Answer 1

Correct Answer: 400

The objective is to determine the mass of water introduced into a closed organ pipe, which shifts its fundamental frequency from 30 Hz to 110 Hz. The pipe's cross-sectional area and the speed of sound in air are provided.

Concept Used:

This problem utilizes the formula for the fundamental frequency of a closed organ pipe (open at one end, closed at the other).

The fundamental frequency (\(f\)) is calculated as:

\[f = \frac{v}{4L}\]

where \(v\) represents the speed of sound in air, and \(L\) is the length of the air column within the pipe.

Introducing water into the pipe reduces the effective length of the air column, thus increasing its fundamental frequency. The volume of added water is derived from the change in air column length and the pipe's cross-sectional area. The mass is then computed 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 air column length (\(L_1\)).

Given the initial fundamental frequency \(f_1 = 30 \, \text{Hz}\) and speed of sound \(v = 330 \, \text{m/s}\), the initial length is:

\[L_1 = \frac{v}{4f_1}\]\[L_1 = \frac{330 \, \text{m/s}}{4 \times 30 \, \text{Hz}} = \frac{330}{120} \, \text{m} = 2.75 \, \text{m}\]

This is the original length of the organ pipe's air column.

Step 2: Calculate the new air column length (\(L_2\)) after water addition.

With the new fundamental frequency \(f_2 = 110 \, \text{Hz}\):

\[L_2 = \frac{v}{4f_2}\]\[L_2 = \frac{330 \, \text{m/s}}{4 \times 110 \, \text{Hz}} = \frac{330}{440} \, \text{m} = \frac{3}{4} \, \text{m} = 0.75 \, \text{m}\]

This is the length of the air column remaining after water has been poured in.

Step 3: Determine the height of the water column.

The reduction in air column length corresponds to the height of the water column (\(h_{water}\)).

\[h_{water} = L_1 - L_2 = 2.75 \, \text{m} - 0.75 \, \text{m} = 2.0 \, \text{m}\]

Step 4: Calculate the volume of the water.

The cross-sectional area is \(A = 2 \, \text{cm}^2\). Convert the water height to centimeters.

\[h_{water} = 2.0 \, \text{m} = 200 \, \text{cm}\]

The volume of 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.

Using the density of water \(\rho_{water} = 1 \, \text{g/cm}^3\):

\[\text{Mass} = \text{Volume} \times \text{Density}\]\[m_{water} = 400 \, \text{cm}^3 \times 1 \, \text{g/cm}^3 = 400 \, \text{g}\]

The mass of water added to the organ pipe is 400 g.