/*! 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 102 A piano tuner hears a beat every... [FREE SOLUTION] | 91Ó°ÊÓ

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A piano tuner hears a beat every 2.00 s when listening to a 264.0 - \(\mathrm{Hz}\) tuning fork and a single piano string. What are the two possible frequencies of the string?

Short Answer

Expert verified
The two possible frequencies of the piano string are \(264.5 Hz\) (higher than the tuning fork frequency) and \(263.5 Hz\) (lower than the tuning fork frequency).

Step by step solution

01

Write down the given information

We are given the following: - Beat frequency = 0.5 Hz - Frequency of tuning fork = 264 Hz Now, we need to find the two possible frequencies (f1 and f2) of the piano string.
02

Use the beat frequency formula to determine the two possible frequencies

Using the beat frequency formula, we can write: \( Beat\ frequency = |f1 - f2| \) We know one of the frequencies - the tuning fork frequency (f2) is 264 Hz. We also know the beat frequency is 0.5 Hz. Let's assume that the piano string has either a higher frequency f1 (case 1) or a lower frequency f1 (case 2) than the tuning fork frequency (f2). Case 1: Piano string frequency (f1) is higher than the tuning fork frequency (f2) \( Beat\ frequency = f1 - f2 \) Solving for f1, we get: \( f1 = Beat\ frequency + f2 \) \( f1 = 0.5 Hz + 264 Hz \) \( f1 = 264.5 Hz \) Case 2: Piano string frequency (f1) is lower than the tuning fork frequency (f2) \( Beat\ frequency = f2 - f1 \) Solving for f1, we get: \( f1 = f2 - Beat\ frequency \) \( f1 = 264 Hz - 0.5 Hz \) \( f1 = 263.5 Hz \)
03

Write down the two possible frequencies

Now we have determined the two possible frequencies of the piano string: 1. If the piano string frequency (f1) is higher than the tuning fork frequency (f2), then f1 = 264.5 Hz. 2. If the piano string frequency (f1) is lower than the tuning fork frequency (f2), then f1 = 263.5 Hz.

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Key Concepts

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

Tuning Fork
A tuning fork is a small, U-shaped piece of metal that vibrates to produce a specific musical note when struck. It is used widely in music and scientific applications because of its ability to generate a pure tone of constant frequency, often ensuring accurate pitch calibration. When a tuning fork is struck, it vibrates at its natural frequency, causing nearby air particles to oscillate and produce sound waves that travel to the listener’s ear. This makes it an excellent tool for tuning musical instruments like pianos. The frequency of the tuning fork mentioned in this exercise is 264 Hz.
This means it produces sound waves that vibrate 264 times per second, which forms the baseline or reference frequency to which the piano string is compared. By listening to the sound from both the tuning fork and the piano string simultaneously, differences in frequency become evident.
Frequency Difference
The frequency difference between two sounds can be detected through what is known as beat frequency. When two sound waves of slightly different frequencies are played together, they interfere with each other. This interference causes the perceived sound to oscillate between loud and soft at a rate equal to the difference in their frequencies.
The number of beats per second heard is referred to as the beat frequency. In our exercise, the beat frequency heard is 0.5 Hz, meaning the sound intensities cycle through loud-soft transitions twice every second. This auditory effect is key in determining the frequency of the unknown sound, in this case, the piano string by offering insights on how close its frequency is to the fixed frequency of the tuning fork.
The beat frequency formula is expressed as:
  • \( \text{Beat frequency} = |f_1 - f_2| \)
where \( f_1 \) and \( f_2 \) are the frequencies of the two sound sources. Here, one of them is the tuning fork with a known frequency.
Piano String Frequency
The frequency of a piano string determines how it sounds when played, with higher frequencies producing higher-pitched notes. In the exercise, the piano tuner's task is to match the piano string frequency as closely as possible to the tuning fork frequency.
Using the beat frequency, we deduced two possibilities for the piano string frequency:
  • If the string frequency is higher than the tuning fork, it can be calculated as:
\[ f_1 = f_2 + \text{Beat frequency} = 264 \text{ Hz} + 0.5 \text{ Hz} = 264.5 \text{ Hz}\]
  • If the string frequency is lower than the tuning fork, it represents:
\[ f_1 = f_2 - \text{Beat frequency} = 264 \text{ Hz} - 0.5 \text{ Hz} = 263.5 \text{ Hz}\]These calculations reflect the two potential pitches the piano string could be tuned to, allowing for adjustment in either direction depending on whether the piano string is tuned sharp or flat relative to the tuning fork.

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Most popular questions from this chapter

A 512-Hz tuning fork is struck and placed next to a tube with a movable piston, creating a tube with a variable length. The piston is slid down the pipe and resonance is reached when the piston is \(115.50 \mathrm{cm}\) from the open end. The next resonance is reached when the piston is \(82.50 \mathrm{cm}\) from the open end. (a) What is the speed of sound in the tube? (b) How far from the open end will the piston cause the next mode of resonance?

What is the minimum speed at which a source must travel toward you for you to be able to hear that its frequency is Doppler shifted? That is, what speed produces a shift of \(0.300 \%\) on a day when the speed of sound is 331 \(\mathrm{m} / \mathrm{s} ?\)

A string \(\left(\mu=0.006 \frac{\mathrm{kg}}{\mathrm{m}}, L=1.50 \mathrm{m}\right)\) is fixed at both ends and is under a tension of 155 N. It oscillates in the \(n=10\) mode and produces sound. A tuning fork is ringing nearby, producing a beat frequency of \(23.76 \mathrm{Hz}\). (a) What is the frequency of the sound from the string? (b) What is the frequency of the tuning fork if the tuning fork frequency is lower? (c) What should be the tension of the string for the beat frequency to be zero?

A sound wave is modeled with the wave function \(\Delta P=1.20 \mathrm{Pa} \sin \left(k x-6.28 \times 10^{4} \mathrm{s}^{-1} t\right)\) and the sound wave travels in air at a speed of \(v=343.00 \mathrm{m} / \mathrm{s}\). (a) What is the wave number of the sound wave? (b) What is the value for \(\Delta P(3.00 \mathrm{m}, 20.00 \mathrm{s}) ?\)

A person has a hearing threshold 10 dB above normal at \(100 \mathrm{Hz}\) and \(50 \mathrm{dB}\) above normal at \(4000 \mathrm{Hz}\). How much more intense must a \(100-\mathrm{Hz}\) tone be than a 4000 - \(\mathrm{Hz}\) tone if they are both barely audible to this person?

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