The speed of sound in air is 344 m>s at room temperature. The lowest frequency of a large organ pipe is 30 s-1 and the highest frequency of a piccolo is 1.5 * 104 s-1. Determine the difference in wavelength between these two sounds.

Short Answer

Expert verified
The wavelength difference is \( \lambda_{\text{organ}} - \lambda_{\text{piccolo}} \approx11.48 \text{ m} - 0.02293 \text{ m} = 11.45707 \text{ m} \).

Step by step solution

01

Understanding the Problem

We need to calculate the difference in wavelength between two sounds of different frequencies using the speed of sound in air. The speed of sound is given as 344 m/s. The frequency of the organ pipe is 30 s-1, and the frequency of the piccolo is 15,000 s-1 (1.5 * 10^4 s-1).
02

Calculating Wavelength of the Organ Pipe Sound

Use the formula for speed (v) and frequency (f) related to wavelength (\( \lambda \)): \( v = \lambda f \). Solve for \( \lambda \) to get \( \lambda = \frac{v}{f} \). Since the frequency of the organ pipe is 30 s-1, the wavelength for the organ pipe sound is \( \lambda_{\text{organ}} = \frac{344 \text{ m/s}}{30 \text{ s}^{-1}} \).
03

Calculating Wavelength of the Piccolo Sound

Similarly, for the piccolo sound, calculate the wavelength using the frequency of the piccolo, which is 15,000 s-1: \( \lambda_{\text{piccolo}} = \frac{344 \text{ m/s}}{15000 \text{ s}^{-1}} \).
04

Determining the Difference in Wavelength

Subtract the wavelength of the piccolo sound from the wavelength of the organ pipe sound to find the difference in wavelength: \( \Delta\lambda = \lambda_{\text{organ}} - \lambda_{\text{piccolo}} \). Calculate this difference using the results from Steps 2 and 3.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Speed of Sound
Starting with the basics, the speed of sound refers to how fast sound waves travel through a medium. This medium can be a solid, liquid, or gas. The speed at which sound travels can be affected by various factors, including temperature, humidity, and air pressure. In our exercise, we're considering the speed of sound in air at room temperature, which is given as 344 meters per second (m/s).

This value is not arbitrary; it is a determined speed at a given temperature of approximately 20°C. It's crucial to recognize that sound travels faster in warmer air and slower in colder air because the molecules move more vigorously at higher temperatures. Understanding this parameter is fundamental because it aids in solving problems related to sound waves, such as calculating wavelengths and frequencies.
Sound Frequency
Frequency, in the context of sound waves, refers to the number of vibrations or cycles per second. It is measured in Hertz (Hz), where one Hertz is equivalent to one cycle per second. In our textbook exercise, we analyze two distinct frequencies: a large organ pipe with a low frequency of 30 Hz and a piccolo which produces a much higher frequency of 15,000 Hz (1.5 * 10^4 Hz).

The frequency of a sound wave is directly linked to our perception of pitch; a higher frequency corresponds to a higher pitch and vice versa. This concept is important as it connects with the wavelength and helps determine the pitch of sounds, which is a central theme in music, acoustics, and audio engineering fields. When calculating wavelength differences, understanding the frequency component of sound waves allows us to visualize the physical length between the peaks of each wave for different sounds.
Wavelength of Sound Waves
The wavelength of a sound wave is the physical distance from one point of a wave to the identical point on the next wave. Think of it as measuring the distance from one wave's crest to the next crest. It is represented by the Greek letter lambda (\( \text{\lambda} \text{\). In a formulaic relationship, wavelength is inversely related to frequency and can be computed if the speed of sound in the medium is known, using the equation \( \text{\lambda} = \frac{v}{f} \). In simpler terms, as frequency increases, the wavelength of the sound decreases, and as frequency decreases, the wavelength increases.}

Related to the exercise, we observed this inverse relationship when calculating the wavelength for two sounds with differing frequencies. A lower frequency sound, such as from the organ pipe, has a longer wavelength, whereas a high-frequency sound from the piccolo has a shorter wavelength. By computing these wavelengths, we can find out how far apart sound waves are for two drastically different musical instruments. This becomes particularly useful when studying the physics of music and designing instruments and concert halls for optimal acoustics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A proton in a linear accelerator has a de Broglie wavelength of 122pm. What is the speed of the proton?

The wave functions for the 1s and 2s orbitals are as follows: 1s c = (1>p)1>2 (1/a3>2 0 ) exp (-r>a0), 2s c = (1>32p)1>2 (1/a3>2 0 ) (2-r>a0) exp(-r>a0) where a0 is a constant (a0 = 53 pm) and r is the distance from the nucleus. Use a spreadsheet to make a plot of each of these wave functions for values of r ranging from 0 pm to 200 pm. Describe the differences in the plots and identify the node in the 2s wave function.

Which set of quantum numbers cannot occur together to specify an orbital? a. n = 2, l = 1, ml = -1 b. n = 3, l = 2, ml = 0 c. n = 3, l = 3, ml = 2 d. n = 4, l = 3, ml = 0

Determine whether an interference pattern is observed on the other side of the slits in each experiment. a. An electro n beam is aimed at two closely spaced slits. The beam is attenuated (made dimmer) to produce only 1 electron per minute. b. An electron beam is aimed at two closely spaced slits. A light beam is placed at each slit to determine when an electron goes through the slit. c. A high-intensity light beam is aimed at two closely spaced slits. d. A gun is fired at a solid wall containing two closely spaced slits. (Will the bullets that pass through the slits form an interference pattern on the other side of the solid wall?)

In a technique used for surface analysis called auger electron spectroscopy (AES), electrons are accelerated toward a metal surface. These electrons cause the emissions of secondary electrons called auger electrons from the metal surface. The kinetic energy of the auger electrons depends on the composition of the surface. The presence of oxygen atoms on the surface results in auger electrons with a kinetic energy of approximately 506 eV. What is the de Broglie wavelength of one of these electrons?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free