91Ó°ÊÓ

Sound waves are essentially oscillations of pressure through a medium, such as air, water, or solid substances. The pressure variation in a sound wave can be described using a sinusoidal equation. In the exercise, the equation provided is: \( \Delta P = 0.0035 \sin(0.38\pi x - 1350\pi t) \). In this form, \( \Delta P \) represents the deviation from the ambient pressure at a point in the medium, indicating how this pressure varies over time and space.

Understanding sound waves begins with recognizing these pressure variations since they determine how energy is transported through a medium. Each peak and trough in the wave represents either a compression or a rarefaction of particles in the medium. As the wave progresses, particles move slightly, altering the local density, which leads to pressure changes that your ears perceive as sound.

The concept of wavelength is fundamental in understanding sound waves. Wavelength (\( \lambda \)) is the distance between consecutive points of similar phase in the wave, such as from crest to crest.

In the exercise, the wave number \( k \) is calculated as \( 0.38\pi \). It is related to the wavelength by the formula:

• \( k = \frac{2\pi}{\lambda} \)

By rearranging this, we derive the wavelength:
\[ \lambda = \frac{2\pi}{0.38\pi} \approx 5.263 \text{ meters} \]

Wavelength is crucial because it gives insight into the size of one complete cycle of the wave. Different wavelengths result in different pitches when it comes to sound: longer wavelengths produce lower-pitched sounds, while shorter wavelengths produce higher-pitched sounds.

Frequency is another vital property of sound waves, symbolized as \( f \), and represents the number of cycles the wave completes in one second. In the equation identified in our problem, we are given the angular frequency \( \omega = 1350\pi \). Frequency can be calculated using:

• \( \omega = 2\pi f \)

Solving for \( f \), we have:
\[ f = \frac{1350\pi}{2\pi} = 675 \text{ Hz} \]

The unit hertz (Hz) corresponds to cycles per second. The frequency directly correlates with the pitch of sound; higher frequencies imply higher-pitched notes, while lower frequencies indicate deeper sounds. This attribute helps in various applications from music to communication technologies.

The speed of a sound wave is a result of how quickly the pressure changes move through the medium. It is determined using the relationship:

• \( v = f \lambda \)

This relates frequency and wavelength with the speed (\( v \)) of the wave.

By substituting the calculated frequency \( 675 \text{ Hz} \) and wavelength \( 5.263 \text{ meters} \) from prior calculations:
\[ v = 675 \times 5.263 \approx 3550 \text{ m/s} \]

This speed can vary significantly based on the medium through which the sound is traveling. Sound moves faster in denser mediums, such as solids, than in gases. Understanding wave speed is crucial in areas like acoustics, engineering, and meteorology.

Displacement amplitude describes the maximum distance that particles in the medium move from their resting positions as the wave passes. In essence, it measures the extent of oscillation in the medium and is a key factor in determining the loudness of the sound.

For the given problem, displacement amplitude \( s_m \) is calculated using:

• \( s_m = \frac{\Delta P_m}{\rho v} \)

Given \( \Delta P_m = 0.0035 \) Pa, \( \rho = 2.3 \times 10^3 \text{ kg/m}^3 \), and wave speed \( 3550 \text{ m/s} \):
\[ s_m = \frac{0.0035}{2.3 \times 10^3 \times 3550} \approx 4.31 \times 10^{-10} \text{ meters} \]

Despite being a tiny value, displacement amplitude significantly affects the perceived volume of sound; larger amplitudes equate to louder sounds. This concept is vital for sound quality control in numerous audio applications and technologies.