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Two identical loudspeakers separated by distance \(d\) emit \(170 \mathrm{Hz}\) sound waves along the \(x\) -axis. As you walk along the axis, away from the speakers, you don't hear anything even though both speakers are on. What are three possible values for d? Assume a sound speed of \(340 \mathrm{m} / \mathrm{s}\)

Short Answer

Expert verified
The three possible separations of the speakers are 1m, 3m, and 5m.

Step by step solution

01

Calculate the Wavelength

First, determine the wavelength of the sound wave. To do this, rearrange the wave speed equation \(v = fλ\) to solve for the wavelength \(λ\). This gives \(λ = v / f\). Substituting in the given expression for the speed of sound \(v = 340 \, m/s\) and the frequency \(f = 170 \, Hz\), we find that the wavelength \(λ = 340 \, m/s / 170 \, Hz = 2 \, m\).
02

Determine Possible Distances

Destructive interference occurs when the path difference of the two sound waves is \(λ/2, 3λ/2, 5λ/2\), etc (odd multiples of \(λ/2\)). Since \(λ = 2\,m\), the three possible values for the distance \(d\) separating the speakers are \( λ/2 = 2\,m / 2 = 1\,m\), \(3λ/2 = 3 \cdot 2\,m / 2 = 3\,m\), and \(5λ/2 = 5 \cdot 2\,m / 2 = 5\,m\).

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Key Concepts

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

Destructive interference
When two waves meet and interact, they can interference with each other, leading to either a louder sound or complete silence. Destructive interference is a neat phenomenon where two sound waves cancel each other out, resulting in no sound being heard. This happens when the waves are out of phase, meaning their peaks and troughs align oppositely.

In layman's terms, imagine one wave's peak is hitting the other wave's trough—this causes them to cancel each other. For destructive interference to occur, the difference in the path length (the distance each wave has traveled) must be a multiple of half the wavelength, like \(\lambda/2, 3\lambda/2, 5\lambda/2\), and so on.
  • \(\lambda/2\) aligns peak to trough, cancelling each other out.
  • Odd multiples like \(3\lambda/2\) or \(5\lambda/2\) continue this pattern.

It's fascinating because even though both speakers are active, you don't hear anything because of this interference!
Sound waves
Sound waves are vibrations that travel through air, water, or solids. They are longitudinal waves, meaning the particles in the wave oscillate back and forth in the same direction the wave travels. This movement is what our ears pick up as sound.

Key characteristics of sound waves include:
  • Frequency: The number of complete wave cycles passing a point per second, measured in Hertz (Hz). Higher frequency means a higher pitch.
  • Amplitude: Relates to loudness; bigger amplitude equals louder sound.
  • Wavelength \(\lambda\): The distance between consecutive points of equivalent phase on the wave, such as crest to crest or trough to trough.

Sound waves can interfere with each other, altering what we hear. When the crest of one wave meets the trough of another, like in the phenomenon of destructive interference, they can cancel each other out, leading to silence.
Wavelength calculation
To understand any wave, it's crucial to calculate its wavelength. This is fairly straightforward. The wavelength is the distance a wave travels during one cycle.

Here's how you compute it:
  • Use the wave speed equation: \( v = f\lambda \), where \( v \) is the velocity of the sound wave, \( f \) is its frequency, and \( \lambda \) the wavelength.
  • Rearrange to find \( \lambda \): \( \lambda = v / f \).

For sound waves traveling at 340 meters per second with a frequency of 170 Hz, inserting these into the equation gives: \[ \lambda = \frac{340 \ m/s}{170 \ Hz} = 2 \ m \]
This result means the sound wave covers 2 meters in each complete cycle. Understanding and calculating wavelength is fundamental for determining patterns of interference, such as figuring out distances in destructive interference scenarios!

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Most popular questions from this chapter

A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of \(39.1 \mathrm{cm} .\) The room is slightly warm, so the speed of sound is \(344 \mathrm{m} / \mathrm{s}\). If the string's linear density is \(0.600 \mathrm{g} / \mathrm{m}\) and the tension is \(150 \mathrm{N},\) how long is the vibrating section of the violin string?

Two loudspeakers \(42.0 \mathrm{m}\) apart and facing each other emit identical 115 Hz sinusoidal sound waves in a room where the sound speed is \(345 \mathrm{m} / \mathrm{s} .\) Susan is walking along a line between the speakers. As she walks, she finds herself moving through loud and quiet spots. If Susan stands \(19.5 \mathrm{m}\) from one speaker, is she standing at a quiet spot or a loud spot?

The width of a particular microwave oven is exactly right to support a standing-wave mode. Measurements of the temperature across the oven show that there are cold spots at each edge of the oven and at three spots in between. The wavelength of the microwaves is \(12 \mathrm{cm} .\) How wide is the oven?

A \(40-\mathrm{cm}\) -long tube has a \(40-\mathrm{cm}-\) long insert that can be pulled in and out, as shown in Figure \(\mathrm{P} 16.61 .\) A vibrating tuning fork is held next to the tube. As the insert is slowly pulled out, the sound from the tuning fork creates standing waves in the tube when the total length \(L\) is \(42.5 \mathrm{cm}, 56.7 \mathrm{cm},\) and \(70.9 \mathrm{cm} .\) What is the frequency of the tuning fork? The air temperature is \(20^{\circ} \mathrm{C}\).

The speed of sound in room temperature \(\left(20^{\circ} \mathrm{C}\right)\) air is \(343 \mathrm{m} / \mathrm{s} ;\) in room temperature helium, it is \(1010 \mathrm{m} / \mathrm{s} .\) The fundamental frequency of an open-closed tube is 315 Hz when the tube is filled with air. What is the fundamental frequency if the air is replaced with helium?

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