Analysis of the problem:

In this problem, we are given a tube ABCDE which is open at A and D. A source of sound is placed in front of A and the frequency of the source can be varied from 2000 Hz to 4000 Hz. We need to find the frequencies at which a detector placed in front of D receives a maximum intensity.

Understanding the phenomenon of sound waves:

Sound waves are longitudinal waves that require a medium to propagate. In a tube like ABCDE, the sound waves can travel in both directions. When the sound waves reach the open ends A and D, they can either reflect or transmit.

Conditions for constructive interference:

To find the frequencies at which a detector placed in front of D receives a maximum intensity, we need to consider the conditions for constructive interference. Constructive interference occurs when the path difference between the two waves is equal to an integral multiple of the wavelength.

Resonance in an open tube:

When a sound wave reaches an open end of a tube, it reflects back and undergoes a phase change of 180 degrees. This means that the path difference between the incident wave and the reflected wave is equal to half a wavelength.

Resonance condition:

For constructive interference to occur at the open end D, the path difference between the incident wave and the reflected wave should be equal to an integral multiple of the wavelength. Mathematically, this can be expressed as:

Path difference = n * λ/2

Where n is an integer representing the number of half wavelengths.

Calculating the frequencies:

To find the frequencies at which a detector placed in front of D receives a maximum intensity, we need to calculate the wavelengths corresponding to the given frequency range (2000 Hz to 4000 Hz) and then find the frequencies that satisfy the resonance condition.

We can use the formula v = λ * f, where v is the speed of sound and f is the frequency.

Let's assume the speed of sound in the tube is v. For the fundamental frequency (n = 1), the wavelength λ1 is given by λ1 = 2L, where L is the length of the tube.

For the first overtone (n = 2), the wavelength λ2 is given by λ2 = L.

Similarly, for the second overtone (n = 3), the wavelength λ3 is given by λ3 = 2/3 * L.

Using the formula v = λ * f, we can calculate the frequencies corresponding to these wavelengths.

Results:

The frequencies at which a detector placed in front of D receives a maximum intensity are:

- For the fundamental frequency (n = 1): f1 = v / (2L)
- For the first overtone (n = 2): f2 = v / L
- For the second overtone (n = 3): f3 = 3v / (2L)

Conclusion:

In this problem, we have analyzed the phenomenon of sound waves in a tube and derived the frequencies at which a detector placed in front of D receives a maximum intensity. By considering the conditions for constructive interference and the resonance condition for an open tube, we have calculated the frequencies corresponding to the given wavelength range. These frequencies can be used to set the detector to receive the maximum intensity of sound.

Figure kha hai?