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The right-most key on a piano produces a sound wave that has a frequency of \(4185.6 \mathrm{~Hz}\). Assuming that the speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\), find the corresponding wavelength.

Short Answer

Expert verified
The wavelength is approximately 0.082 meters.

Step by step solution

01

Understand the Relationship Between Frequency and Wavelength

The speed of a wave is related to its frequency and wavelength via the formula: \( v = f \times \lambda \), where \( v \) is the speed of the wave, \( f \) is the frequency, and \( \lambda \) is the wavelength. To find the wavelength, rearrange the formula to \( \lambda = \frac{v}{f} \).
02

Identify Given Data

From the exercise, note the given values: the frequency \( f = 4185.6\, \text{Hz} \) and the speed of sound \( v = 343\, \text{m/s} \).
03

Plug Values into the Wavelength Formula

Insert the given values into the rearranged formula: \[ \lambda = \frac{343\, \text{m/s}}{4185.6\, \text{Hz}}. \]
04

Calculate the Wavelength

Perform the division to calculate \( \lambda \): \[ \lambda = \frac{343}{4185.6} \approx 0.0819 \text{ m}. \]
05

Interpret the Result

The calculated value \( \lambda \approx 0.0819 \text{ m} \) means that the wavelength of the sound wave corresponding to the right-most key on a piano is approximately 0.082 meters.

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Key Concepts

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

Understanding Frequency
Frequency is a fundamental concept in wave physics that refers to the number of cycles a wave undergoes in a given unit of time. It is usually measured in hertz (Hz), where one hertz is equivalent to one cycle per second.
To visualize frequency, imagine the waves generated when you pluck a guitar string. Each oscillation, or cycle, of the string produces a wave, and the frequency is the count of these oscillations per second.
  • High-frequency waves have more cycles per second and typically correspond to higher-pitched sounds.
  • Low-frequency waves have fewer cycles per second and are often perceived as lower-pitched sounds.
In the context of the piano example, the sound wave generated has a high frequency of 4185.6 Hz, indicating a high-pitched note. Such frequency values are common for the uppermost keys of a piano, which produce very sharp tones. Remember, the relationship between a wave's frequency and its other properties, such as wavelength and speed, are crucial for understanding wave behavior.
Exploring Wavelength
Wavelength is the distance between consecutive points of a wave in phase, such as peaks or troughs. It is usually expressed in meters. In simple terms, wavelength is the length of one complete wave cycle.
  • Longer wavelengths often correspond to lower frequency waves and thus lower-pitched sounds.
  • Shorter wavelengths align with higher frequency waves, producing higher-pitched sounds.
To determine wavelength, use the formula: \[ \lambda = \frac{v}{f} \]where \( \lambda \) denotes the wavelength, \( v \) is the speed of sound, and \( f \) is the frequency.
Applying this formula to the piano scenario: \( \lambda = \frac{343\, \text{m/s}}{4185.6\, \text{Hz}} = 0.0819 \text{ m} \), showing that the sound wave has a wavelength of approximately 0.082 meters.
Understanding wavelength is vital not just for physics, but also in music, engineering, and telecommunications, where precise wave manipulations are frequently required.
The Speed of Sound
The speed of sound is an important concept in acoustics and wave physics. It refers to how fast sound travels through a medium, typically air, and is affected by factors like temperature, humidity, and pressure. For air at room temperature, the speed of sound is usually taken to be about 343 meters per second (m/s).
  • In colder temperatures, the speed of sound decreases because air molecules move less energetically.
  • In warmer temperatures, the speed increases due to the higher energy of the air molecules.
In our exercise, the speed of sound plays a critical role in calculating wavelength. Knowing the speed allows us to determine how far the wave travels during one complete cycle, offering insight into the size and behavior of the wave itself.
This concept has practical applications in various fields, like aviation and meteorology, where precise calculations of sound speed can be crucial to ensure accuracy and safety. Moreover, understanding the speed of sound helps musicians and audio engineers adjust settings to obtain desired acoustic effects.

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Most popular questions from this chapter

A rocket, starting from rest, travels straight up with an acceleration of \(58.0 \mathrm{~m} / \mathrm{s}^{2}\). When the rocket is at a height of \(562 \mathrm{~m}\), it produces sound that eventually reaches a ground-based monitoring station directly below. The sound is emitted uniformly in all directions. The monitoring station measures a sound intensity \(I .\) Later, the station measures an intensity \(\frac{1}{3} I\). Assuming that the speed of sound is \(343 \mathrm{~m} / \mathrm{s},\) find the time that has elapsed between the two measurements.

The displacement (in meters) of a wave is given according to \(y=0.26 \sin (\pi t-3.7 \pi x)\) where \(t\) is in seconds and \(x\) is in meters, (a) Is the wave traveling in the \(+x\) or \(-x\) direction? (b) What is the displacement \(y\) when \(t=38 \mathrm{~s}\) and \(x=13 \mathrm{~m} ?\)

Two identical rifles are shot at the same time, and the sound intensity level is \(80.0 \mathrm{~dB}\). What would be the sound intensity level if only one rifle were shot? (Hint: The answer is not \(40.0 \mathrm{dB.}\) )

Concept Questions A uniform rope of mass \(m\) and length \(L\) is hanging straight down from the ceiling. (a) Is the tension in the rope greater near the top or near the bottom of the rope, or is the tension the same everywhere along the rope? Why? (b) A small-amplitude transverse wave is sent up the rope from the bottom end. Is the speed of the wave greater near the bottom or near the top of the rope, or is the speed of the wave the same everywhere along the rope? Explain. (c) Consider a section of the rope between the bottom end and a point that is a distance \(y\) meters above the bottom. What is the weight of this section? Express your answer in terms of \(m, L, y,\) and \(g\) (the acceleration due to gravity). Problem (a) For the rope described in the Concept Questions, derive an expression that gives the speed of the wave on the rope in terms of the distance \(y\) above the bottom end and the acceleration \(g\) due to gravity. (b) Use the expression that you have derived to calculate the speeds at distances of \(0.50 \mathrm{~m}\) and \(2.0 \mathrm{~m}\) above the bottom end of the rope. Be sure that your answers are consistent with your answer to Concept Question (b).

The tension in a string is \(15 \mathrm{~N},\) and its linear density is \(0.85 \mathrm{~kg} / \mathrm{m} .\) A wave on the string travels toward the \(-x\) direction; it has an amplitude of \(3.6 \mathrm{~cm}\) and a frequency of \(12 \mathrm{~Hz}\) What are the (a) speed and (b) wavelength of the wave? (c) Write down a mathematical expression (like Equation 16.3 or 16.4 ) for the wave, substituting numbers for the variables \(A, f,\) and \(\lambda\)

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