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Two speakers are driven in phase by a common oscillator at \(800 \mathrm{Hz}\) and face each other at a distance of \(1.25 \mathrm{m}\) Locate the points along a line joining the two speakers where relative minima of sound pressure amplitude would be expected. (Use \(v=343 \mathrm{m} / \mathrm{s} .\) )

Short Answer

Expert verified
The points of minima sound pressure amplitude are found at \(0.214375\) m and \(0.643125\) m from the mid-point between the two speakers on both sides.

Step by step solution

01

Definition of Given Quantities

Given: Frequency (\(f\)) = 800 Hz, Velocity (\(v\)) = 343 m/s, and Separation distance between speakers (\(d\)) = 1.25 m. Also note that the path difference for destructive interference is an odd multiple of half wavelengths.
02

Calculate Wavelength

First, we need to calculate the wavelength (\(\lambda\)) of the sound wave. Using the wave equation \(v = f \times \lambda\), we can rearrange to find \(\lambda = v/f\). Substituting the given values, we get \(\lambda = 343/800 = 0.42875\) m.
03

Calculate Path Difference

The path difference for destructive interference is given by \((2n - 1) \times (\lambda/2)\), where \(n\) is a positive integer. We consider each possible value of \(n\) until the resulting path difference exceeds the separation between the speakers. Starting with \(n = 1\), we find path differences of \(0.214375\) m, \(0.643125\) m, \(1.071875\) m. Any higher values of \(n\) would give a path difference greater than the speaker separation, so we stop here.
04

Locate the Points of Minima

The points of relative minima of sound pressure amplitude will occur at distances equal to these path differences on either side of the mid-point between the speakers along the line joining them. So, the minima occur at \(0.214375\) m and \(0.643125\) m away from the mid-point on each side.

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Key Concepts

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

Sound Wave Wavelength
Understanding the wavelength of a sound wave is crucial when studying wave phenomena like interference. In physics, the wavelength, denoted as \(\lambda\), is the distance between two consecutive points that are in phase. These points could represent peaks, troughs, or identical positions on different cycles of a wave.

For sound, the wavelength determines the type of auditory experience one can expect. For example, shorter wavelengths correspond to higher-pitched sounds, while longer wavelengths are associated with lower-pitched sounds. When it comes to interference, the wavelength is even more important because it determines where the points of constructive and destructive interference will occur.

In the solution provided, the sound wave emitted by the speakers has a wavelength of \(0.42875 \text{m}\), calculated by dividing the velocity of sound \(v = 343 \text{m/s}\) by the frequency \(f = 800 \text{Hz}\). Therefore, when examining interference, one must consider this wavelength to accurately predict where the pressure minima or maxima may be.
Path Difference for Interference
The path difference is a fundamental concept in the study of wave interference, particularly when identifying regions of constructive or destructive interference. It refers to the difference in distance travelled by two waves before they meet. In the context of sound waves, destructive interference occurs when two waves of the same frequency and amplitude meet out of phase, canceling each other out and resulting in a lower overall sound amplitude.

In our example, to achieve destructive interference, the path difference must be an odd multiple of half the wavelength, expressed as \( (2n - 1) \times (\lambda/2) \), where \( n \) is an integer. This relationship ensures that the waves meet out of phase, producing pockets of silence at specific locations between the speakers. The ability to calculate these points accurately enables us to predict where these areas of silence – the points of relative minima – will occur, as demonstrated in the original exercise solution.
Sound Wave Frequency
Sound wave frequency, measured in Hertz (Hz), reflects the number of complete wave cycles that pass a point in one second. It is directly correlated to the pitch of the sound we hear – higher frequencies produce higher pitched sounds, while lower frequencies are perceived as deeper sounds.

In the context of interference, frequency plays a critical role as it determines the spacing of the interference fringes. Every sound frequency has a corresponding wavelength that follows the equation \(v = f \times \(\lambda\)\), where \(v\) is the speed of the sound wave. In our exercise, we see a frequency of \(800 \text{Hz}\), which in combination with the speed of sound, allows us to calculate the wavelength and then examine the interference patterns. The frequency is a fundamental aspect of not just music and acoustics, but also in understanding the physical properties of waves and how they interact in different environments.

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Most popular questions from this chapter

The wave function for a standing wave is given in Equation 18.3 as \(y=2 A \sin k x \cos \omega t .\) (a) Rewrite this wave function in terms of the wavelength \(\lambda\) and the wave speed \(v\) of the wave. (b) Write the wave function of the simplest standing wave vibration of a stretched string of length \(L\). (c) Write the wave function for the second harmonic. (d) Generalize these results and write the wave function for the \(n\)th resonance vibration.

The top string of a guitar has a fundamental frequency of \(330 \mathrm{Hz}\) when it is allowed to vibrate as a whole, along all of its 64.0 -cm length from the neck to the bridge. A fret is provided for limiting vibration to just the lower two-thirds of the string. (a) If the string is pressed down at this fret and plucked, what is the new fundamental frequency? (b) What If? The guitarist can play a "natural harmonic" by gently touching the string at the location of this fret and plucking the string at about one sixth of the way along its length from the bridge. What frequency will be heard then?

If two adjacent natural frequencies of an organ pipe are determined to be \(550 \mathrm{Hz}\) and \(650 \mathrm{Hz},\) calculate the fundamental frequency and length of this pipe. (Use \(v=\) \(340 \mathrm{m} / \mathrm{s} .)\)

A cello A-string vibrates in its first normal mode with a frequency of \(220 \mathrm{Hz}\). The vibrating segment is \(70.0 \mathrm{cm}\) long and has a mass of \(1.20 \mathrm{g} .\) (a) Find the tension in the string. (b) Determine the frequency of vibration when the string vibrates in three segments.

Two identical sinusoidal waves with wavelengths of \(3.00 \mathrm{m}\) travel in the same direction at a speed of \(2.00 \mathrm{m} / \mathrm{s} .\) The second wave originates from the same point as the first, but at a later time. Determine the minimum possible time interval between the starting moments of the two waves if the amplitude of the resultant wave is the same as that of each of the two initial waves.

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