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An organ pipe open at one end is vibrating in first overtone and is in resonance with another pipe open at both ends and vibrating in third harmonic. The ratio of length of two pipes is (a) \(3: 8\) (b) \(8: 3\) (c) \(1: 2\) (d) \(4: 1\)

Short Answer

Expert verified
The ratio of the length of the two pipes is 1:2 (option c).

Step by step solution

01

Understanding Pipe Resonance

First, let's understand the resonance conditions for organ pipes. An organ pipe open at one end forms a standing wave with nodes and antinodes at its ends. The first overtone of such a pipe corresponds to the third harmonic. For a pipe open at both ends, the third harmonic implies the presence of two complete waves within the pipe.
02

Formulating the Condition for Pipe 1

For an organ pipe open at one end, the fundamental frequency occurs with a node at the open end and an antinode at the closed end. The first overtone (which is actually the second harmonic for open-open pipes) occurs with two nodes and an antinode in between. The length of the pipe corresponds to three-quarter wavelengths: \( L_1 = \frac{3\lambda_1}{4} \).
03

Formulating the Condition for Pipe 2

For an organ pipe open at both ends, there is an antinode at each end. The third harmonic fits one and a half wavelengths in the length of the pipe: \( L_2 = \frac{3\lambda_2}{2} \).
04

Equating the Frequencies of Both Pipes

Since the two pipes are in resonance, they have the same frequency. The frequency is given by \( f = \frac{v}{\lambda} \), where \( v \) is the speed of sound and \( \lambda \) is the wavelength. Since the frequencies are equal, we say: \( \frac{v}{\lambda_1} = \frac{v}{\lambda_2} \) which implies \( \lambda_1 = \lambda_2 \).
05

Finding the Length Ratio

Substitute \( \lambda_1 = \lambda_2 \) into the expressions for \( L_1 \) and \( L_2 \):\[ L_1 = \frac{3}{4}\lambda, \quad L_2 = \frac{3}{2}\lambda \].Taking the ratio, we have:\[ \frac{L_1}{L_2} = \frac{\frac{3}{4}\lambda}{\frac{3}{2}\lambda} = \frac{3}{4} \times \frac{2}{3} = \frac{1}{2} \].
06

Final Solution Conclusion

The length ratio of the two pipes, \( L_1 : L_2 \), is \( 1:2 \). Hence, the answer is (c) \( 1:2 \).

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Key Concepts

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

Harmonics in Physics
Harmonics are key to understanding why musical instruments, like organ pipes, produce specific sounds and pitches. In physics, harmonics refer to the frequencies at which an object naturally resonates. They are integer multiples of the fundamental frequency, which is the lowest frequency produced by an oscillating system.
When an organ pipe vibrates at different harmonics, it means that the system can support standing waves of different lengths and shapes. For example:
  • The fundamental frequency is the first harmonic and represents the simplest standing wave pattern.
  • The second harmonic is twice the frequency of the fundamental and incorporates more nodes and antinodes.
  • Higher harmonics continue this pattern, allowing for increasingly complex waveforms.
Understanding harmonics helps us predict how an organ pipe will behave under certain acoustic conditions and is essential for designing instruments that need precise control over their musical output.
Standing Waves in Pipes
Standing waves are formed when waves of the same frequency interfere while traveling in opposite directions. In organ pipes, this principle creates distinct patterns called harmonics. In pipes, the waves reflect off the ends, setting up resonance conditions to ensure that standing waves are sustained.
There are two primary types of pipes:
  • Pipes open at one end, where one end of the pipe is closed, and the other is open.
  • Pipes open at both ends, with open ends on both sides of the pipe.
Each type of pipe supports different standing wave patterns:
- For a pipe open at one end, the fundamental pattern is a quarter wavelength long, with a node at the closed end and an antinode at the open end.
- For a pipe open at both ends, the fundamental pattern is half a wavelength long, with antinodes at both ends.
These fundamental patterns and their multiples form the harmonics that allow pipes to produce sound.
Acoustic Resonance Conditions
Resonance occurs when an object vibrates at its natural frequency, amplified by external forces at specific frequencies. In the context of organ pipes, resonance is achieved when standing waves align perfectly with the pipe's physical dimensions.
For pipes:
  • An open-open pipe resonates when the pipe's length matches half the wavelength of the sound wave at a particular harmonic.
  • An open-closed pipe resonates similarly but with a quarter-wavelength pattern for its fundamental frequency.
The resonance condition is crucial for musical consistency and achieving the desired sound. It involves meticulous calculation to ensure that the organ pipes play in harmony with their intended frequencies, whether in isolation or amidst other pipes.
Thus, understanding acoustic resonance conditions is essential for crafting instruments that sound vibrant and harmonious.

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Most popular questions from this chapter

A stretched string of length \(l\) fixed at both ends can sustain stationary waves of wavelength \(\lambda\) given by (a) \(\lambda=2 \ln\) (b) \(\lambda=2 l / n\) (c) \(\lambda=l^{2} / 2 n\) (d) \(\lambda=n^{2} / 2 l\)

When 2 tuning forks (fork 1 and fork 2) are sounded simultaneously, 4 beats \(\mathrm{s}^{-1}\) are heard. Now, some tape is attached on the prong of fork \(2 .\) When the tuning forks are sounded again, 6 beats \(\mathrm{s}^{-1}\) are heard if the frequency of fork 1 is \(200 \mathrm{~Hz}\), then what was the original frequency of fork 2 ? (a) \(196 \mathrm{~Hz}\) (b) \(200 \mathrm{~Hz}\) (c) \(202 \mathrm{~Hz}\) (d) \(204 \mathrm{~Hz}\)

A wire of density \(9 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) is stretched between two clamps \(1 \mathrm{~m}\) apart and is subjected to an extension of \(4.9 \times 10^{-4} \mathrm{~cm} .\) The lowest frequency of transverse vibration in the wire is ( \(Y=9 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}\) ) (a) \(40 \mathrm{~Hz}\) (b) \(35 \mathrm{~Hz}\) (c) \(30 \mathrm{~Hz}\) (d) \(25 \mathrm{~Hz}\)

Two uniform wires are vibrating simultaneously in their fundamental notes. The tension, lengths diameters and the densities of the two wires are in the ratio \(8: 1,36: 35,4: 1\) and \(1: 2\) respectively. If the note of the higher pitch has a frequency \(360 \mathrm{~Hz}\), the number of beats produced per second is (a) 5 (b) 15 (c) 10 (d) 20

A tuning fork of frequency \(480 \mathrm{~Hz}\) produces 10 beats \({ }^{-1}\) when sounded with a vibrating sonometer string. What must have been the frequency of string if slight increase in tension produces fewer beats \(\mathrm{s}^{-1}\) than before? (a) \(490 \mathrm{~Hz}\) (b) \(470 \mathrm{~Hz}\) (c) \(460 \mathrm{~Hz}\) (d) \(480 \mathrm{~Hz}\)

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