Waves

The separation between two successive nodes is not equal to the wavelength , but half of the wavelength . This is because nodes in standing waves are separated by half the wavelength, as seen in the harmonic wave pattern. Therefore, statement (A) is incorrect. The other statements are correct:

• In an open organ pipe, antinodes are indeed formed at both ends.
• In a one-end closed organ pipe, a node is formed at the closed end, and an antinode is formed at the open end.
• For a stretched string, nodes are formed at both ends.
• The separation between the successive node and antinode is , which is a characteristic of standing wave patterns.

The correct option is (A) : The separation between two successive nodes is

Understand the Physics Principles:
The frequency of a stretched string (sonometer wire) vibrating in its fundamental mode is given by: where is the length of the vibrating segment, is the tension in the wire, and is the linear mass density (mass per unit length). Since the tension and linear density are constant for the same wire under the same conditions, the frequency is inversely proportional to the length: Beats occur when two sound waves of slightly different frequencies interfere. The beat frequency is the absolute difference between the two frequencies:

Define Variables and Set Up Equations:
Let be the frequency of the tuning fork. Let cm and cm be the two lengths of the sonometer wire. Let and be the fundamental frequencies of the sonometer wire corresponding to lengths and , respectively. The beat frequency is given as Hz for both lengths. Since , and , we must have . The tuning fork produces 4 beats/sec with both lengths, so: Since , and the tuning fork frequency results in the same beat frequency with both and , must lie between and . If it were outside this range (either or ), the differences and could not both be equal to 4 Hz unless , which contradicts . Therefore, we must have . This leads to: From equations (1) and (2), we can find the difference between and :

Use the Frequency-Length Relationship:
We know that . Therefore: Substitute cm and cm:

Solve for the Frequencies:
Substitute the expression for into equation (3): Now find using equation (3):

Calculate the Tuning Fork Frequency:
Using equation (1): Alternatively, using equation (2): Both methods yield the same result.

The frequency of the tuning fork is 256 Hz. This corresponds to option (D).

Given:

• Speed of tuning fork (source), (moving towards observer)
• Observed frequency,
• Speed of sound,
• Observer is stationary ( )

Step 1: Apply Doppler Effect Formula

When a sound source moves towards a stationary observer, the observed frequency is given by:

where is the actual frequency of the tuning fork.

Step 2: Solve for Actual Frequency ( )

Rearrange the formula to solve for :

Substitute the known values:

Conclusion:

The actual frequency of vibration of the tuning fork is 300 Hz.

Answer:

The fundamental frequency of a string clamped at both ends is given by:

where:
- is the length of the string (1 m),
- is the tension in the string (1 N),
- is the linear mass density of the string ( ).

Let's calculate the fundamental frequency:

The angular frequency is:

The displacement of the string in the fundamental mode can be described as a standing wave with nodes at the ends and an antinode in the middle. The displacement is given by:

where is the amplitude of the wave (1 cm = 0.01 m).

The acceleration is the second time derivative of the displacement:

The maximum acceleration occurs at the antinode (middle of the string, ) where and :

Thus the correct Answer is Option C .

Thus, the maximum acceleration in the string should be .

Step 1: The general equation of a stationary wave is: where:
- is the amplitude,
- is the wave number,
- is the position,
- is the angular frequency.
In the given equation, we have .
Step 2: The wave number is related to the wavelength by: From the given equation, , so the wavelength is: Step 3: The distance between two consecutive nodes is half the wavelength: Thus, the distance between two consecutive nodes is 2 cm.