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Two identical piano wires have a fundamental frequency of \(600 \mathrm{~Hz}\) when kept under the same tension. What fractional increase in the tension of one wire will lead to the occurrence of \(6.0\) beats/s when both wires oscillate simultaneously?

Short Answer

Expert verified
The fractional increase in tension is 0.0404.

Step by step solution

01

Understanding the Beat Frequency

When two waves of slightly different frequencies are played together, beats are produced with a beat frequency given by \[ f_{beat} = |f_1 - f_2|. \]We are told that the beat frequency is 6 beats per second, so \[ |f_1 - f_2| = 6 \].
02

Relating Frequency to Tension

The fundamental frequency \( f \) of a string is related to the tension \( T \) and mass per unit length \( \mu \) by the formula\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \].For the wires with identical tension initially, \( f_1 = f_2 = 600 \, \text{Hz} \). If we increase the tension in one wire, the new frequency \( f_1 \) will be affected.
03

Expressing New Frequency in Terms of Increased Tension

Let the new tension be \( T' = T + \Delta T \). Then the new frequency is given by\[ f_1 = \frac{1}{2L} \sqrt{\frac{T'}{\mu}} = \frac{1}{2L} \sqrt{\frac{T + \Delta T}{\mu}}. \]
04

Approximating Tension Increase Using Beat Frequency

The frequency of the second wire remains \( 600 \, \text{Hz} \).We know from the beat frequency that \[ |f_1 - 600| = 6, \]so either \( f_1 = 606 \, \text{Hz} \) or \( f_1 = 594 \, \text{Hz} \). Given that tension increases to raise frequency,\[ f_1 = 606 \, \text{Hz}. \]
05

Solving for the Fractional Increase in Tension

We have \[ 606 = \frac{1}{2L} \sqrt{\frac{T + \Delta T}{\mu}} \]. \Divide both sides by 600:\[ \frac{606}{600} = \sqrt{\frac{T + \Delta T}{T}}. \]Squaring both sides, we get\[ \left(\frac{606}{600}\right)^2 = \frac{T + \Delta T}{T}. \]Calculate:\[ \left(\frac{606}{600}\right)^2 = 1.02^2 = 1.0404 \].Thus,\[ \frac{T + \Delta T}{T} = 1.0404.\]Therefore, \[ \frac{\Delta T}{T} = 1.0404 - 1 = 0.0404.\] The fractional increase in tension is 0.0404.

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Key Concepts

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

Tension in Strings
When we talk about tension in strings in physics, we refer to the force exerted along the string. This tension is a key factor that affects how strings vibrate. Imagine a piano wire: the tension within it determines not only its pitch but also how it reacts when played. The tighter the string, the higher the frequency of the sound it produces. This concept is crucial in musical instruments that use strings.

In terms of mathematics, the frequency of vibration of a string, denoted as \( f \), is linked to the tension \( T \) in the string by a specific formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]where \( L \) represents the length of the string and \( \mu \) is the mass per unit length.

What's happening inside this formula? The equation shows that as tension increases, the frequency of the sound wave produced also increases, provided mass and length stay constant. So, in our piano wire example, increasing the tension means the wire vibrates at a higher frequency, yielding a higher pitch sound.
Fundamental Frequency
The term **fundamental frequency** refers to the lowest frequency at which a system like a string on a piano vibrates. This frequency is the base one, upon which other harmonics are built. Musical notes are often identified by their fundamental frequencies.

For strings, the fundamental frequency is where it feels most "at home"; it’s the simplest form of vibration for the string. In mathematical terms, it’s calculated using:\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]As seen from this equation, the fundamental frequency increases with higher tension and decreases with more mass or length. If you imagine a guitar or a piano tuner adjusting the tension in their instrument strings, they're effectively changing the fundamental frequencies to get the desired pitch.

When we play two strings of close but different fundamental frequencies together, something interesting happens—beat frequency. It's essential to understand the fundamental frequency to grasp how instruments generate rich, harmonious sounds.
Wave Interference
Wave interference is a phenomenon that occurs when two or more waves superimpose, or lay on top of each other, resulting in a new wave pattern. This is observed in our daily lives and has significant implications in music, particularly with string instruments like the piano.

When two waves of similar frequencies are played together, they interfere, leading to what is known as "beats." This is where wave interference takes center stage in understanding sound phenomena.

**Beat Frequency** - The beat frequency \( f_{beat} \) is given by: \[ f_{beat} = |f_1 - f_2| \] - This formula tells us how the slight difference in frequencies between two waves results in a pulsating sound, or "beats."

This concept helps us understand why a piano might need tuning. If one wire on the piano is slightly out of tune, you might hear irregular beats. By correcting the tension—thus the fundamental frequency—it ensures smooth, clear sound waves with no unwanted interference. Wave interference showcases how precise adjustments make significant impacts in both physics and music.

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Most popular questions from this chapter

A bat is flitting about in a cave, navigating via ultrasonic bleeps. Assume that the sound emission frequency of the bat is \(39000 \mathrm{~Hz}\). During one fast swoop directly toward a flat wall surface, the bat is moving at \(0.025\) times the speed of sound in air. What frequency does the bat hear reflected off the wall?

An ambulance with a siren emitting a whine at \(1600 \mathrm{~Hz}\) overtakes and passes a cyclist pedaling a bike at \(2.44 \mathrm{~m} / \mathrm{s}\). After being passed, the cyclist hears a frequency of \(1590 \mathrm{~Hz}\). How fast is the ambulance moving?

A violin string \(30.0 \mathrm{~cm}\) long with linear density \(0.650 \mathrm{~g} / \mathrm{m}\) is placed near a loudspeaker that is fed by an audio oscillator of variable frequency. It is found that the string is set into oscillation only at the frequencies 880 and \(1320 \mathrm{~Hz}\) as the frequency of the oscillator is varied over the range \(500-1500 \mathrm{~Hz}\). What is the tension in the string?

A stationary motion detector sends sound waves of frequency \(0.150 \mathrm{MHz}\) toward a truck approaching at a speed of \(45,0 \mathrm{~m} / \mathrm{s}\). What is the frequency of the waves reflected back to the detector?

(a) Find the speed of waves on a violin string of mass \(800 \mathrm{mg}\) and length \(22.0 \mathrm{~cm}\) if the fundamental frequency is \(920 \mathrm{~Hz}\) (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string?0

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