Question

The pressure in a traveling sound wave is given by the equation $$ \Delta p=(1.48 \mathrm{~Pa}) \sin [(1.07 \pi \mathrm{rad} / \mathrm{m}) x-(334 \pi \mathrm{rad} / \mathrm{s}) t] $$ Find \((a)\) the pressure amplitude, (b) the frequency, \((c)\) the wavelength, and \((d)\) the speed of the wave.

Short answer

The pressure amplitude is 1.48 Pa, the frequency is 167 Hz, the wavelength is 1.869 m, and the speed of the wave is 312.15 m/s.

Step-by-step solution

Finding the Pressure Amplitude

The pressure amplitude is the constant in front of the sine function, thus it is 1.48 Pa.

Finding the Frequency

The angular frequency \(ω\) is 334π rad/s. According to the formula \(ω = 2πf\), the frequency \(f\) is \(ω / 2π = 334 / 2 = 167 Hz.\)

Determining the Wavelength

The angular wave number \(k\) is 1.07π rad/m. By using the formula \(λ = 2π / k\), we can find the wavelength \(λ = 2 / 1.07 = 1.869 m.\)

Calculating the Speed of the Wave

The speed of the wave \(v\) is determined by dividing the angular frequency by the angular wave number, hence \(v = ω / k = 334 / 1.07 = 312.15 m/s.\)

Key concepts

Pressure Amplitude

When dealing with sound waves, the **pressure amplitude** is crucial to understanding just how intense the wave is. The pressure amplitude refers to the maximum change in pressure caused by the sound wave as it travels through a medium.

In the given equation, \( \Delta p=(1.48 \, \text{Pa}) \sin [(1.07 \pi \, \text{rad/m}) x-(334\pi \,\text{rad/s}) t] \), the pressure amplitude is the constant that appears in front of the sinusoidal function, which shows maximum extents of pressure change. In this equation, it is particularly given as \(1.48 \, \text{Pa}\).

This value indicates the highest deviation from the normal air pressure that the sound wave causes in its medium. A higher pressure amplitude corresponds to a louder sound, as more significant pressure changes are perceived as higher intensity by the human ear.

Frequency Calculation

The **frequency** of a sound wave is a measure of how many times the wave oscillates or cycles each second. It’s crucial for determining the pitch of the sound: higher frequencies mean a higher-pitched sound.

To calculate frequency from the given wave equation, we need to focus on the angular frequency component, which is represented by \(\omega\) in the formula. Here, \(\omega = 334\pi \,\text{rad/s}\). We use the formula \(\omega = 2\pi f\) to solve for \(f\), which represents frequency in Hertz (Hz).

By rearranging the formula, we find \( f = \frac{\omega}{2\pi} = \frac{334\pi}{2\pi} = 167 \, \text{Hz} \). This frequency tells us that the sound wave completes 167 cycles every second, which is a typical frequency range for sounds that humans can hear.

Wavelength Determination

The **wavelength** is the distance between consecutive points of equivalent phase in the wave, like crest to crest or trough to trough. It’s key in determining how different sound waves interact with their environments.

In our sound wave equation, the angular wave number \(k\) is \(1.07\pi \, "\text{rad/m}\); it relates directly to the wavelength. To find the wavelength \(\lambda\), use the formula \(\lambda = \frac{2\pi}{k}\).

Plugging in our values, we calculate \( \lambda = \frac{2\pi}{1.07\pi} = \frac{2}{1.07} \approx 1.869 \, \text{m} \). The wavelength captures the length of one complete cycle of the wave in meters. Shorter wavelengths tend to correspond to higher frequency sounds, while longer wavelengths are lower in frequency.

Wave Speed

The **wave speed** is essential for understanding how quickly a sound wave travels through a medium. In simple terms, it’s how fast the wave crests are moving.

For this sound wave, the wave speed \(v\) can be found using the relationship between its angular frequency \(\omega\) and its angular wave number \(k\). The formula to find speed is \(v = \frac{\omega}{k}\).

Using the provided values, \(\omega = 334\pi \,\text{rad/s}\) and \(k = 1.07\pi \,\text{rad/m}\), we calculate the speed \(v = \frac{334\pi}{1.07\pi} = \frac{334}{1.07} \approx 312.15 \, \text{m/s}\).

This speed tells us how fast the wave is propagating through the medium, which in this case is likely air. Wave speed can vary based on the medium's properties like temperature and density.