91Ó°ÊÓ

In a sound wave, the pressure amplitude represents the maximum change in pressure from the ambient atmospheric pressure due to the wave. It essentially measures how intense or loud the sound is perceived.

In our given sound wave equation, the pressure amplitude is the coefficient of the sine function. This means it's the number that multiplies the sine part of the equation itself, which indicates the peak deviation from the standard pressure. In this particular equation, the pressure amplitude is directly given as \(1.50 \, \text{Pa}\). This value tells us the greatest extent, in pascals, to which the pressure changes above and below the normal atmospheric pressure as the wave passes through a medium.

Angular frequency is a crucial concept for understanding how fast the sound wave oscillates over time. It is denoted by \( \omega \) and is measured in radians per second (s\(^{-1}\)).

In the context of the wave equation, angular frequency shows up as a part of the term that multiplies time \( t \). Specifically, it was identified as \( 315 \, \text{s}^{-1} \) in the equation \( \Delta p = (1.50 \, \text{Pa}) \sin \pi [0.900 \, \text{m}^{-1} x - 315 \, \text{s}^{-1} t ] \). This value represents how quickly the sine wave oscillates as time progresses.

• Angular frequency is linked to the physical frequency of the wave (\( f \)) by the formula \( \omega = 2 \pi f \).
• In this exercise, using this relationship helps derive the frequency of the wave which was calculated as approximately \( 50.2 \, \text{Hz} \).

The wavelength of a sound wave, denoted by \( \lambda \), is the physical distance over which the wave's shape repeats itself. Long wavelengths correspond to lower frequencies, and vice versa.

Wavelength is naturally tied to the wave number \( k \), which is derived through the relationship \( k = \frac{2\pi}{\lambda} \). In this particular problem, the wave number was provided as \( 0.900 \, \text{m}^{-1} \). Solving for the wavelength using \( \lambda = \frac{2\pi}{k} \), we find it to be approximately \( 6.98 \, \text{m} \).

Having the wavelength gives us insight into the size of the sound wave in spatial terms. It tells us how far the sound travels before the pattern begins to repeat.

The speed of a sound wave, represented by \( v \), describes how quickly the sound wave propagates through a medium. It's an essential parameter because it tells us how fast the sound reaches our ears or detectors.

The formula connecting speed with frequency and wavelength is \( v = f \lambda \). From the exercise, the calculated frequency \( f \) is approximately \( 50.2 \, \text{Hz} \), and the wavelength \( \lambda \) is approximately \( 6.98 \, \text{m} \). Plugging these into the formula, we find the speed of the wave to be about \( 350.4 \, \text{m/s} \).

• This speed could vary depending on factors like temperature and medium (air, water, etc.).
• In most exercises involving ideal conditions, the calculated speed aligns closely with the standard speed of sound in air, which is approximately \( 343 \, \text{m/s} \) at room temperature.