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Two identical taut strings under the same tension \(F\) produce a note of the same fundamental frequency \(f_{0}\) . The tension in one of them is now increased by a very small amount \(\Delta F\) . (a) If they are played together in their fundamental, show that the frequency of the beat produced is \(f_{\text { keet }}=f_{0}(\Delta F / 2 F)\) . (b) Two identical violin strings, when in tune and stretched with the same tension, have a fundamental frequency of 440.0 \(\mathrm{Hz}\) . One of the strings is retuned by increasing its tension. When this is done, 1.5 beats per second are heard when both strings are plucked simultaneously at their centers. By what percentage was the string tension changed?

Short Answer

Expert verified
The change in string tension was approximately 0.682%.

Step by step solution

01

Understand the Relationship of Frequency with Tension

The frequency for a string under tension is given by the formula \( f = \frac{1}{2L} \sqrt{\frac{F}{\mu}} \), where \( L \) is the length of the string and \( \mu \) is the linear mass density. For identical strings, we focus on the squared relationship between frequency and tension: \( f \propto \sqrt{F} \).
02

Calculate the Changed Frequency

Considering two strings with tensions \( F \) and \( F + \Delta F \), the frequencies are \( f_0 \) and \( f \). From the relation \( f \propto \sqrt{F} \), we write:\[\frac{f}{f_0} = \sqrt{\frac{F + \Delta F}{F}}.\] Given \( \Delta F \) is small, we can use the binomial approximation, \( \sqrt{1+x} \approx 1 + \frac{x}{2} \), so:\[\frac{f}{f_0} \approx 1 + \frac{\Delta F}{2F}.\] Thus, \( f \approx f_0 + f_0 \left(\frac{\Delta F}{2F}\right) \).
03

Determine the Beat Frequency

The beat frequency \( f_{\text{beat}} \) is the absolute difference between the two frequencies \( f \) and \( f_0 \), hence:\[f_{\text{beat}} = |f - f_0| = f_0 \left(\frac{\Delta F}{2F}\right).\] This confirms part (a) of the problem.
04

Apply Given Data for Beat Frequency

From part (b), knowing \( f_{\text{beat}} = 1.5 \) Hz and \( f_0 = 440.0 \) Hz, substitute into the beat frequency formula:\[1.5 = 440 \times \left(\frac{\Delta F}{2F}\right).\] Solve for \( \frac{\Delta F}{F} \):\[\frac{\Delta F}{2F} = \frac{1.5}{440} \implies \frac{\Delta F}{F} = \frac{3}{440}.\]
05

Calculate Percentage Change in Tension

To find the percentage change in tension, multiply \( \frac{\Delta F}{F} \) by 100%:\[\text{Percentage change} = \left(\frac{3}{440}\right) \times 100 \approx 0.682\%.\]

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Key Concepts

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

Frequency and Tension Relationship
The relationship between frequency and tension in a stringed instrument is an essential part of understanding how sounds are produced. For a string under tension, the frequency is primarily defined by the formula \( f = \frac{1}{2L} \sqrt{\frac{F}{\mu}} \), where \( f \) is the frequency, \( F \) is the tension, \( L \) is the length of the string, and \( \mu \) is the linear mass density. For two identical strings, this simplifies to show that the frequency is proportional to the square root of the tension: \( f \propto \sqrt{F} \).

When the tension in a string is increased, the frequency of the sound produced also increases. This squared relationship is why even small changes in tension can significantly affect frequency. Understanding this principle is critical for tuning stringed instruments and achieving the expected sound.
Beat Frequency
Beat frequency describes the phenomenon where two frequencies close in value produce an audible beating effect. This outcome occurs when two slightly different frequencies interfere with each other. The beat frequency \( f_{\text{beat}} \) is calculated using the absolute difference between the two frequencies involved: \[ f_{\text{beat}} = |f_2 - f_1| \]

In the exercise discussed, two strings are played together at their fundamental frequencies. One string's tension is slightly increased, altering its frequency slightly. The beat frequency becomes the difference in their frequencies. This principle is crucial for musicians, as it helps in tuning instruments by identifying whether strings produce the same frequency when played together. By adjusting tension to minimize beats, one can achieve perfect tuning.
Fundamental Frequency
The fundamental frequency, often referred to as the first harmonic, is the lowest frequency produced by a vibrating string. For an ideal string fixed at both ends, this is the frequency at which the string naturally vibrates when plucked or struck. The term is applied generally to describe the principal note in a complex sound composed of multiple harmonics.

For any given string, the fundamental frequency depends on its tension, density, and length. In the exercise, both strings initially have the same fundamental frequency, 440 Hz. It is the note that primarily determines the pitch of the sound heard. Adjusting tension changes this fundamental frequency, which further affects the sound quality and pitch perceived.
Binomial Approximation
The binomial approximation is a mathematical method used to simplify calculations involving powers of expressions close to one. The formula \( \sqrt{1+x} \approx 1 + \frac{x}{2} \) is commonly employed when \( x \) is small. This method allows us to approximate complex expressions in the exercise.

In the context of the discussed problem, when the tension \( F \) is changed slightly (\( \Delta F \) is small), we can approximate the change in frequency caused by this tension adjustment. The approximation gives a quicker pathway to find the resultant frequency change when the change in tension is not significant. This simplification aids in solving practical problems like determining beat frequencies without delving into detailed, cumbersome calculations.

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Most popular questions from this chapter

Longitudinal Waves in Different Fluids. (a) A longitudinal wave propagating in a water-filled pipe has intensity \(3.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\) and frequency 3400 \(\mathrm{Hz}\) . Find the amplitude \(A\) and wavelength \(\lambda\) of the wave. Water has density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) and bulk modulus \(2.18 \times 10^{9} \mathrm{Pa}\) . Water has density 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) at pressure \(1.00 \times 10^{5} \mathrm{Pa}\) and density 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) , what will be the amplitude \(A\) and wavelength \(\lambda\) of a longitudinal wave with the same intensity and frequency as in part (a)? (c) In which fluid is the amplitude larger, water or air? What is the ratio of the two amplitudes? Why is this ratio so different from 1.00\(?\)

The shock-wave cone created by the space shuttle at one instant during its reentry into the atmosphere makes an angle of \(58.0^{\circ}\) with its direction of motion. The speed of sound at this altitude is 331 \(\mathrm{m} / \mathrm{s}\) (a) What is the Mach number of the shuttle at this instant, and (b) how fast (in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{mi} / \mathrm{h} )\) is it traveling relative to the atmosphere? (c) What would be its Mach number and the angle of its shock-wave cone if it flew at the same speed but at low altitude where the speed of sound is 344 \(\mathrm{m} / \mathrm{s} ?\)

(a) Show that the fractional change in the speed of sound \((d v / v)\) due to a very small temperature change \(d T\) is given by \(d v / v=\frac{1}{2} d T / T .\) (Hint: Start with Eq. \(16.10 .\) (b) The speed of sound in air at \(20^{\circ} \mathrm{C}\) is found to be 344 \(\mathrm{m} / \mathrm{s}\) . Use the result in part (a) to find the change in the speed of sound for a \(1.0^{\circ} \mathrm{C}\) change in air temperature.

A woman stands at rest in front of a large, smooth wall. She holds a vibrating tuning fork of frequency \(f_{0}\) directly in front of her (between her and the wall). (a) The woman now runs toward the wall with speed \(v_{\mathrm{W}}\) . She detects beats due to the interference between the sound waves reaching her directly from the fork and those reaching her after being reflected from the wall. How many beats per second will she detect? (Note: If the beat frequency is too large, the woman may have to use some instrumentation other than her ears to detect and count the beats.) (b) If the woman instead runs away from the wall, holding the tuning fork at her back so it is between her and the wall, how many beats per second will she detect?

A certain pipe produces a fundamental frequency of 262 \(\mathrm{Hz}\) in air. (a) If the pipe is filled with helium at the same temperature, what fundamental frequency does it produce? (The molar mass of air is 28.8 \(\mathrm{g} / \mathrm{mol}\) , and the molar mass of helium is 4.00 \(\mathrm{g} / \mathrm{mol}\) . (b) Does your answer to part (a) depend on whether the pipe is open or stopped? Why or why not?

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