/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 If \(\lambda_{1}, \lambda_{2}\),... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 69

If \(\lambda_{1}, \lambda_{2}\), and \(\lambda_{3}\) are the wavelengths of the waves giving resonance with the fundamental, first and second overtones, respectively, of a closed organ pipe, then the ratio of wavelengths \(\lambda_{1}: \lambda_{2}: \lambda_{3}\) is (A) \(1: 2: 3\) (B) \(1: \frac{1}{3}: \frac{1}{5}\) (C) \(1: 3: 5\) (D) \(5: 3: 1\)

Short Answer

Expert verified
The ratio of wavelengths for the fundamental, first, and second overtones in a closed organ pipe is \(\lambda_{1}:\lambda_{2}:\lambda_{3} = 1:\frac{1}{3}:\frac{1}{5}\).

Step by step solution

01

Find the formula for wavelength

Using the given formula for the resonance frequencies and the relation between wavelength and frequency, we can derive the formula for the wavelength of the fundamental and overtones: \[\lambda = \frac{4L}{(2n+1)}\]
02

Determine the wavelengths of the fundamental and overtones

For the fundamental, n = 0: \[\lambda_1 = \frac{4L}{(2\cdot 0 + 1)} = \frac{4L}{1}= 4L\] For the first overtone, n = 1: \[\lambda_2 = \frac{4L}{(2\cdot 1 + 1)} = \frac{4L}{3}\] For the second overtone, n = 2: \[\lambda_3 = \frac{4L}{(2\cdot 2 + 1)} = \frac{4L}{5}\]
03

Determine the ratio of wavelengths

Now that we have the formulas for the wavelengths of the fundamental and overtones, we can find their ratio: \[\lambda_1:\lambda_2:\lambda_3 = 4L:\frac{4L}{3}:\frac{4L}{5}\] To find the ratio without including the common factor 4L, we can multiply each term by 15 (the least common multiple of the denominators 1, 3, and 5): \[15 \cdot (4L:\frac{4L}{3}:\frac{4L}{5}) = 60L:20L:12L\] Finally, simplify the ratio by dividing each term by their greatest common divisor, 4L: \[\frac{60L}{4L}:\frac{20L}{4L}:\frac{12L}{4L} = 15:5:3\] Substituting these simplified wavelengths back to the original ratio, we get: \[\lambda_1:\lambda_2:\lambda_3 = 1:\frac{1}{3}:\frac{1}{5}\] Thus, the correct answer is (B) \(1:\frac{1}{3}:\frac{1}{5}\).

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 91影视!

Key Concepts

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

Wavelength
In the realm of physics, particularly in wave mechanics, the concept of wavelength is pivotal when discussing sound waves. Wavelength (\( \lambda \)) measures the spatial period of the wave鈥攖he distance between two identical points that are in phase, such as peak to peak or trough to trough.

For a closed organ pipe, the wavelength plays a crucial role in defining the sounds that resonate within it. Closed organ pipes produce sound through standing waves, and each wave corresponds to a specific note depending on the wavelength. To find the wavelength for a closed organ pipe, we use the formula: \[ \lambda = \frac{4L}{(2n+1)} \] where \( L \) is the length of the pipe and \( n \) represents the harmonic number.

This formula highlights that only odd harmonics resonate in a closed organ pipe, and the wavelength is inversely proportional to the harmonic number (n). Therefore, for the fundamental harmonic (n=0), the wavelength is longest, and each subsequent overtone has a shorter wavelength.
Fundamental Frequency
The fundamental frequency is the lowest frequency produced by a vibrating object, and it is often considered the "base" frequency for the sound produced. In a closed organ pipe, the fundamental frequency reflects the largest wavelength possible that fits standing wave conditions inside the pipe.

The formula for determining the fundamental frequency of a closed organ pipe is derived from the speed of sound and the wavelength:\[ f_1 = \frac{v}{\lambda_1} \]where \( f_1 \) is the fundamental frequency, \( v \) is the speed of sound in air, and \( \lambda_1 \) is the wavelength of the fundamental frequency.

When an organ pipe is closed at one end, it supports standing waves where there is a node (point of zero amplitude) at the closed end and an antinode (point of maximum amplitude) at the open end. The length of the pipe is an odd multiple of quarter wavelengths, which directly impacts the fundamental frequency produced. The fundamental is key, as it sets the stage for all overtones, which are multiples of this base frequency.
Overtones
Overtones are frequencies higher than the fundamental frequency, and in musical acoustics, they are crucial in shaping the sound's timbre or color. In a closed organ pipe, overtones occur at frequencies above the fundamental frequency due to higher harmonics of the standing wave.

Unlike open pipes, where all harmonics are present, closed organ pipes only support odd harmonics (like the 1st, 3rd, 5th, etc.). This is because of how the wave reflects off the closed end, resulting in only certain patterns being sustainable. Overtones impact the richness and character of the sound, making the pipe's acoustic signature unique.

The first overtone occurs when \( n=1 \),\[ \lambda_2 = \frac{4L}{3} \]and the second overtone at \( n=2 \),\[ \lambda_3 = \frac{4L}{5} \]. Similar to the fundamental, the frequencies of these overtones can be calculated using the speed of sound, enhancing the diversity and complexity of the sound a closed organ pipe can produce.

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

Following are equations of four waves: (i) \(y_{1}=a \sin \omega\left(t-\frac{x}{v}\right)\) (ii) \(y_{2}=a \sin \omega\left(t+\frac{x}{v}\right)\) (iii) \(z_{1}=a \sin \omega\left(t-\frac{x}{v}\right)\) (iv) \(z_{2}=a \cos \omega\left(t+\frac{x}{v}\right)\) Which of the following statement is correct? (A) On superposition of waves (i) and (iii), a traveling wave having amplitude \(a\) will be formed. (B) Superposition of waves (ii) and (iii) is not possible. (C) On superposition of (i) and (ii), a stationary wave having amplitude \(a \sqrt{2}\) will be formed. (D) On superposition of (iii) and (iv), a transverse stationary wave will be formed.

A siren creates a sound level of \(60 \mathrm{~dB}\) at a location of \(500 \mathrm{~m}\) from the speaker. The siren is powered by a battery that delivers a total energy of \(1 \mathrm{~kJ}\). The efficiency of siren is \(30 \%\). The total time for which the siren sound is (A) \(95 \mathrm{~s}\) (B) \(95.5 \mathrm{~s}\) (C) \(96 \mathrm{~s}\) (D) \(96.5 \mathrm{~s}\)

A simple harmonic motion (SHM) has an amplitude \(A\) and time period \(T\). The time required by it to travel from \(x=A\) to \(x=A / 2\) is (A) \(T / 6\) (B) \(T / 4\) (C) \(T / 3\) (D) \(T / 2\)

A train moves towards a stationary observer with speed \(34 \mathrm{~m} / \mathrm{s}\). The train sounds a whistle and its frequency registered is \(f_{1}\). If the train's speed is reduced to \(17 \mathrm{~m} / \mathrm{s}\), the frequency registered is \(f_{2}\). If the speed of sound is \(340 \mathrm{~m} / \mathrm{s}\) then the ratio \(\frac{f_{1}}{f_{2}}\) is (A) \(\frac{18}{19}\) (B) \(\frac{1}{2}\) (C) 2 (D) \(\frac{19}{18}\)

The power of sound from the speaker of a radio is \(20 \mathrm{~mW}\). By turning the knob of volume control, the power of sound is increased to \(400 \mathrm{~mW}\). The power increase in \(\mathrm{dB}\) as compared to the original power is \(\left(\log _{10} 2=0.3\right)\) (A) \(1.3 \mathrm{~dB}\) (B) \(3.1 \mathrm{~dB}\) (C) \(13 \mathrm{~dB}\) (D) \(30.1 \mathrm{~dB}\)
See all solutions

Recommended explanations on Physics Textbooks

Further Mechanics and Thermal Physics

Read Explanation

Classical Mechanics

Read Explanation

Magnetism

Read Explanation

Energy Physics

Read Explanation

Atoms and Radioactivity

Read Explanation

Medical Physics

Read Explanation
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.