(A) A circular diaphragm 60 cm in diameter oscillates at a frequency of 25 kHz as an underwater

(A) A circular diaphragm 60 cm in diameter oscillates at a...

(A) A circular diaphragm $60 \mathrm{~cm}$ in diameter oscillates at a frequency of $25 \mathrm{kHz}$ as an underwater source of sound used for submarine detection. Far from the source, the sound intensity is distributed as the diffraction pattern of a circular hole whose diameter equals that of the diaphragm. Take the speed of sound in water to be $1450 \mathrm{~m} / \mathrm{s}$ and find the angle between the normal to the diaphragm and a line from the diaphragm to the first minimum. (b) Is there such minimum for a source having an (audible) freguency of $1.0 \mathrm{kHz}$ ?

(A) A circular diaphragm $60 \mathrm{~cm}$ in diameter oscillates at a frequency of $25 \mathrm{kHz}$ as an underwater source of sound used for submarine detection. Far from the source, the sound intensity is distributed as the diffraction pattern of a circular hole whose diameter equals that of the diaphragm. Take the speed of sound in water to be $1450 \mathrm{~m} / \mathrm{s}$ and find the angle between the normal to the diaphragm and a line from the diaphragm to the first minimum.
(b) Is there such minimum for a source having an (audible) freguency of $1.0 \mathrm{kHz}$ ?

Key Concepts

Circular Aperture Diffraction

This concept refers to the phenomenon where waves passing through a circular opening spread out or diffract to form a pattern characterized by a central maximum and successive minima and maxima. The pattern is typically described by Bessel functions, and the first minimum occurs at an angle approximately given by 1.22?/D, where ? is the wavelength and D is the diameter of the aperture. This relation is crucial for understanding resolution limits in systems using circular apertures.

Wavelength, Frequency, and Speed Relation

The relationship between wavelength (?), frequency (f), and propagation speed (v) is given by the equation ? = v/f. In the context of sound waves, this formula allows us to compute the wavelength in any medium when the speed of sound and the frequency are known. This is essential for linking the physical dimensions of an aperture to the resulting diffraction pattern.

Angular Distribution of Diffraction Minima

This concept involves determining the angles at which minima in intensity occur in a diffraction pattern. For a circular aperture, the first minimum is found at an angle where the product of the aperture diameter and the sine of the angle is about 1.22 times the wavelength. This angular distribution is a key measure in applications such as acoustic signaling and imaging where beam spread or resolution is important.

Sound Propagation in Different Media

Understanding how sound travels in various media is fundamental to acoustics. The speed of sound, which can vary significantly from one medium to another (for example, from air to water), determines the wavelength for a given frequency. This variation affects the diffraction pattern produced by a sound source and is particularly important in applications like underwater acoustics and sonar.

Transcript

0:00 Hi there.

00:01 So for this problem, we have a circular diagram of 60 centimeters in diameter.

00:09 So we are given the diameter, 60 centimeters that we know in meters is equal to 0 .6 meters.

00:18 And it oscillates at a frequency that is also given, and that frequency is equal to 25 kilohertz.

00:33 And as an underwater source of sound used for submarine detection.

00:39 Now, for the source, the stone intensity is distributed as a diffraction pattern of a circular hole, whose diameter equals that of a diagram.

00:52 Now, we need to take the speed of the sun of the water, the speed of the sun in water, it's equal to 1 ,450 meters per second.

01:10 And we need to find the algorithm between the normal to the diagram and a line from the diagram to the first minimum.

01:17 So that is the first part of this problem.

01:22 So we need to find the angle theta.

01:26 Now for this, we use the equation that follows, that theta is equal to the sign of minus 1 of the product between 1 .22 the wavelength over the distance d.

01:46 Now in this case we know that the distance d we will have that is 0 .6 meters and the wavelength can be reading in terms of the frequency because we know that that is the speed of the sun in water over the frequency so substituting that in here we will have that that is the sign of minus one of 1 .22 times the speed of the sun in water over the frequency this over the distance d now we just need to simply substitute all of those values, the diameter, the frequency, and the speed of the sound in water.

02:37 So we will find that this is the sign of minus 1 .22.

02:44 The speed of the sound underwater, we know that that is 1 ,450 meters per second.

02:56 The frequency is given, and that is 25 kilohertz.

03:01 We know that kilos means tens to the 3 hearths...

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