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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
A 2.01% increase in tension leads to 6 beats/s.

Step by step solution

01

Understand the Problem

We have two identical piano wires, both vibrating at a fundamental frequency of \(600 \text{ Hz}\). We need to find the fractional increase in the tension applied to one of these wires such that when both wires are oscillated, they produce \(6\) beats per second.
02

Apply Beats Formula

When two frequencies are close to each other, they produce beats. The number of beats per second (beat frequency) is the absolute difference between the two frequencies: \( f_{1} - f_{2} = 6 \text{ Hz} \). Hence, if the first wire stays at \(600 \text{ Hz}\), the second wire must change to \(600 + 6 = 606 \text{ Hz}\).
03

Use Frequency-Tension Relationship

The frequency of a vibrating wire is given by the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension, and \( \mu \) is the mass per unit length, same for both wires. Since both wires are identical, the change in frequency is due to the change in tension. We have: \( \frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1}} \).
04

Calculate Fractional Change in Tension

Given \( f_2 = 606 \text{ Hz} \) and \( f_1 = 600 \text{ Hz} \), we find \( \frac{606}{600} = 1.01 \). So, \( 1.01 = \sqrt{\frac{T_2}{T_1}} \). Solving for \( \frac{T_2}{T_1} \), we square both sides: \( 1.01^2 \approx 1.0201 \). Consequently, the fractional increase in tension is \( 0.0201 \) or \( 2.01\% \).

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Key Concepts

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

Frequency
Frequency is a fundamental concept in physics, especially when dealing with waves and oscillations. It measures how many cycles of a wave occur in one second and is typically measured in Hertz (Hz). In the case of musical instruments, frequency determines the pitch we hear.
To illustrate, imagine plucking a piano wire. The wire will vibrate, and this vibration produces sound waves. Each complete vibration or cycle of the wire forms a wave, contributing to the frequency. A higher frequency means the wire vibrates more rapidly, producing a higher sound pitch, while a lower frequency results in a deeper sound pitch.
  • Piano wires, like those in our problem, often have a specific fundamental frequency. This frequency is the lowest possible frequency of vibration for the wire.
  • In the provided exercise, both piano wires initially have a fundamental frequency of 600 Hz.
  • Changes in tension can affect this frequency, as seen in this problem, where altering the tension caused the frequency of one wire to slightly differ.
Understanding how frequency relates to wave mechanics and tension helps explain why two notes may sound discordant or harmonious.
Tension in Strings
Tension in a string is a key factor that influences how a string vibrates and, consequently, the frequency of the sound it produces. Increasing or decreasing tension will alter the sound the string emits when vibrated.
Consider a tightrope walker: the tension in the rope affects how they balance and how the rope behaves. Similarly, in our scenario with piano wires, the tension applied biologically determines their vibration characteristics.
  • Piano strings are under considerable tension, but minute adjustments can significantly affect the sound.
  • In our problem, one wire's tension was increased to achieve a frequency that allowed the production of 6 beats per second.
  • The relationship between tension and frequency is mathematically expressed in the formula: \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \) where \( f \) is frequency, \( L \) is the string length, \( T \) is tension, and \( \mu \) is mass per unit length.
This tension adjustment led to a change in frequency, creating a beat phenomenon due to the slight frequency difference between the two wires.
Wave Mechanics
Wave mechanics involves understanding how waves behave and interact. In wave mechanics, waves can interfere with one another, leading to fascinating phenomena like beats, standing waves, and resonance.

Beats occur when two waves of slightly different frequencies overlap, leading to alternating constructive and destructive interference. When the waves align perfectly (constructive interference), they amplify each other, which we perceive as a louder sound. When they are out of sync (destructive interference), they diminish each other.
  • This is why, in our exercise, the 600 Hz and 606 Hz frequencies produce 6 beats per second; these beats are a direct consequence of the frequency difference.
  • When tuning musical instruments, players use beats to finely adjust the instrument until no beats are heard, indicating that the strings have matching frequencies.
Understanding wave mechanics is pivotal in many areas of physics, from acoustics to optics, and is fundamental in exploring how we perceive and interact with sound.

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Most popular questions from this chapter

The water level in a vertical glass tube \(1.00 \mathrm{~m}\) long can be adjusted to any position in the tube. A tuning fork vibrating at \(686 \mathrm{~Hz}\) is held just over the open top end of the tube, to set up a standing wave of sound in the air-filled top portion of the tube. (That air filled top portion acts as a tube with one end closed and the other end open.) (a) For how many different positions of the water level will sound from the fork set up resonance in the tube's air-filled portion, which acts as a pipe with one end closed (by the water) and the other end open? What are the (b) least and (c) second least water heights in the tube for resonance to occur?

The sound intensity is \(0.0080 \mathrm{~W} / \mathrm{m}^{2}\) at a distance of \(10 \mathrm{~m}\) from an isotropic point source of sound. (a) What is the power of the source? (b) What is the sound intensity \(5.0 \mathrm{~m}\) from the source? (c) What is the sound level \(10 \mathrm{~m}\) from the source? 96 Four sound waves are to be sent through the same tube of air, in the same direction: $$ \begin{array}{l} s_{1}(x, t)=(9.00 \mathrm{~nm}) \cos (2 \pi x-700 \pi t) \\ s_{2}(x, t)=(9.00 \mathrm{~nm}) \cos (2 \pi x-700 \pi t+0.7 \pi) \\ s_{3}(x, t)=(9.00 \mathrm{~nm}) \cos (2 \pi x-700 \pi t+\pi) \\ s_{4}(x, t)=(9.00 \mathrm{~nm}) \cos (2 \pi x-700 \pi t+1.7 \pi) \end{array} $$ What is the amplitude of the resultant wave? (Hint: Use a phasor diagram to simplify the problem.)

A certain sound source is increased in sound level by \(30.0 \mathrm{~dB}\). By what multiple is (a) its intensity increased and (b) its pressure amplitude increased?

A point source emits \(30.0\) W of sound isotropically. A small microphone intercepts the sound in an area of \(0.750 \mathrm{~cm}^{2}, 200 \mathrm{~m}\) from the source. Calculate (a) the sound intensity there and (b) the power intercepted by the microphone.

A point source that is stationary on an \(x\) axis emits a sinusoidal sound wave at a frequency of \(686 \mathrm{~Hz}\) and speed \(343 \mathrm{~m} / \mathrm{s}\). The wave travels radially outward from the source, causing air molecules to oscillate radially inward and outward. Let us define a wavefront as a line that connects points where the air molecules have the maximum, radially outward displacement. At any given instant, the wavefronts are concentric circles that are centered on the source. (a) Along \(x\), what is the adjacent wavefront separation? Next, the source moves along \(x\) at a speed of \(110 \mathrm{~m} / \mathrm{s}\). Along \(x\), what are the wavefront separations (b) in front of and (c) behind the source?

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