91Ó°ÊÓ

A sinusoidal sound wave moves at \(343 \mathrm{~m} / \mathrm{s}\) through air in the positive direction of an \(x\) axis. At one instant, air molecule \(A\) is at its maximum displacement in the negative direction of the axis while air molecule \(B\) is at its equilibrium position. The separation between those molecules is \(15.0 \mathrm{~cm}\), and the molecules between \(A\) and \(B\) have intermediate displacements in the negative direction of the axis. (a) What is the frequency of the sound wave? In a similar arrangement, for a different sinusoidal sound wave, air molecule \(C\) is at its maximum displacement in the positive direction while molecule \(D\) is at its maximum displacement in the negative direction. The separation between the molecules is again \(15.0 \mathrm{~cm}\), and the molecules between \(C\) and \(D\) have intermediate displacements. (b) What is the frequency of the sound wave?

Step by step solution

01

Understanding the Situation for Molecules A and B

Molecule A is at its maximum displacement in the negative direction, and molecule B is at its equilibrium position. This means A and B are separated by a quarter of the wavelength because A is at the crest (or trough) and B is at the equilibrium position.

02

Relate Separation to Wavelength for A and B

The separation of 15.0 cm between molecules A and B is one quarter of the wavelength: \[\text{Separation} = \frac{\lambda}{4}\] Thus,\[ \lambda = 4 \times 0.15 \text{ m} = 0.60 \text{ m}.\]

03

Calculate Frequency for Situation (a)

Using the wave speed equation, where speed \(v = 343\ \text{m/s}\), we have:\[v = f \lambda \] Substituting known values,\[ 343 = f \times 0.60.\] Solving for \(f\),\[f = \frac{343}{0.60} \approx 571.67 \text{ Hz}. \]

04

Understanding the Situation for Molecules C and D

In this case, molecules C and D are at maximum displacement in opposite directions, indicating they are half a wavelength apart.

05

Relate Separation to Wavelength for C and D

The separation of 15.0 cm between molecules C and D is half of the wavelength:\[\text{Separation} = \frac{\lambda}{2}\]Thus,\[ \lambda = 2 \times 0.15 \text{ m} = 0.30 \text{ m}. \]

06

Calculate Frequency for Situation (b)

Using the wave speed equation again:\[v = 343 = f \lambda \] Substituting known values for \(\lambda\),\[ 343 = f \times 0.30 \] Solving for \(f\),\[f = \frac{343}{0.30} \approx 1143.33 \text{ Hz}. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Speed

Wave speed is the speed at which a wave travels through a medium. It is a key factor in determining how quickly sound can move from one place to another.
For sound waves traveling through air, the wave speed depends on the temperature and the properties of the air. In this particular case, the wave speed is given as \(343 \text{m/s}\).
The wave speed can be calculated using the formula:

• \( v = f \lambda \)

where \( v \) is the wave speed, \( f \) is the frequency of the wave, and \( \lambda \) is the wavelength. This formula shows the relationship between speed, frequency, and wavelength, demonstrating how they interplay to describe wave movement.

Wavelength

Wavelength is the distance between consecutive repeating points in a wave pattern, like the distance from crest to crest or trough to trough.
It is often symbolized by the Greek letter \( \lambda \). For sound waves, knowing the wavelength helps us understand other properties of the wave, like its frequency.
In the problems given, separating molecules A and B at a quarter wavelength and molecules C and D at a half wavelength gives vital clues about the sound's characteristics.
When the separation is known, wavelengths can be deduced easily:

• For A and B: \(\lambda = 4 \times 0.15 \text{m} = 0.60 \text{m}\).
• For C and D: \(\lambda = 2 \times 0.15 \text{m} = 0.30 \text{m}\).

Understanding how the physical separation between molecules relates to wavelength allows us to solve for frequency using the wave speed formula.

Sinusoidal Waves

Sinusoidal waves are the most common type encountered in physics and engineering. They describe a smooth, periodic oscillation.
This wave form can be visually represented by a sine or cosine curve. Sound waves are often sinusoidal, moving through air as chains of compressions and rarefactions.
In a sinusoidal wave, every point along the wave has a displacement from equilibrium that follows a sine or cosine function. This means the characteristics of the wave, such as peaks and troughs, repeat in a regular pattern.
In the exercise, both parts (a) and (b) involve sinusoidal waves. Molecules at maximum displacement highlight crests and troughs, helping in visualizing how the wave moves through the air.

Displacement of Molecules

Displacement in sound refers to how far the molecules of air move from their original position as the wave passes through them.
Maximum displacement means that the molecule is at the crest or trough of the sinusoidal wave. At equilibrium, molecules are at their resting state with zero displacement.
In the case study, molecule A at maximum displacement in the negative direction shows it's on a trough, while molecule B at equilibrium indicates it's at the midpoint of the wave cycle.
For molecules C and D being at opposite maximum displacements (positive and negative), it indicates they are positioned at a crest and a trough, respectively. Thus, understanding displacement helps us visualize wave phases and predict molecular positions along a wave.