Sound Wave

The voltmeter is connected in parallel with the 200 resistor. The equivalent resistance of this parallel combination is:

The total resistance of the circuit is:

Using the voltage division rule, the voltage across the 100 resistor is given as:

Substitute the given values:

Simplify by dividing through by 4:

Take the reciprocal:

Multiply through by 100:

Rearrange:

Simplify by cross-multiplying:

Expand terms:

Cancel on both sides:

Solve for :

Final Answer: The resistance of the voltmeter is:

200 .

Consider an open pipe and a closed pipe of the same length . The fundamental frequency of an open pipe is given by , where is the speed of sound. The frequency of the nth overtone for an open pipe is .

For a closed pipe, the fundamental frequency is . The frequency of the nth overtone for a closed pipe is .

To find the seventh overtone frequencies:

• Open Pipe: The frequency of the seventh overtone ( ) is:
  .
• Closed Pipe: The seventh overtone corresponds to the 15th harmonic ( ):
  .

Given , substitute the expressions for and :

Solving this gives:

• Cross-multiply:

Recognize an oversight; given the range is , verify simplifies correctly to align with physical constraints and range.
Correct calculation: isolate and verify positive solution .

The problem asks for the difference in frequencies between two open organ pipes of different lengths, each resonating at a specific harmonic. We are given the lengths of the pipes, the harmonic numbers, and the velocity of sound in air.

Concept Used:

The solution is based on the formula for the frequencies of harmonics in an open organ pipe. An open organ pipe is open at both ends, which means it has antinodes at both ends. As a result, it can support all integer harmonics (both odd and even).

The frequency of the -th harmonic ( ) for an open organ pipe of length is given by:

where:

• is the harmonic number ( ).
• is the velocity of sound in the medium (air).
• is the length of the pipe.

Step-by-Step Solution:

Step 1: List the given values and convert them to SI units.

• Length of the first pipe, .
• Harmonic number for the first pipe, .
• Length of the second pipe, .
• Harmonic number for the second pipe, .
• Velocity of sound in air, .

Step 2: Calculate the resonant frequency of the first organ pipe ( ).

Using the formula for the -th harmonic with and :

Step 3: Calculate the resonant frequency of the second organ pipe ( ).

Using the formula for the -th harmonic with and :

Step 4: Calculate the difference between the two frequencies.

The difference in frequencies is .

The difference of frequencies for the given modes is 740 Hz.

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 ( ) is given by:

where is the speed of sound in air, and 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 ( ).

Step-by-Step Solution:

Step 1: Calculate the initial length of the air column ( ).

The initial fundamental frequency is , and the speed of sound is . Using the formula for the fundamental frequency:

This is the initial length of the organ pipe.

Step 2: Calculate the new length of the air column ( ) after adding water.

The new fundamental frequency is .

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 ( ) is the difference between the initial and final lengths of the air column.

Step 4: Calculate the volume of the water poured into the pipe.

The cross-sectional area is given as . It is convenient to convert the height of the water to centimeters.

The volume of the water is:

Final Computation & Result:

Step 5: Calculate the mass of the water.

The density of water is .

The amount of water poured into the organ tube is 400 g.

The general wave equation for a sound wave is ,

where: - is the amplitude, - is the wave number, and - is the angular frequency.

The wave velocity is given by the formula:

Thus, the correct answer is option (1).