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Chapter 12: Problem 30

A violin is tuned by adjusting the tension in the strings. Brian's A string is tuned to a slightly lower frequency than Jennifer's, which is correctly tuncd to \(440.0 \mathrm{Hz}\) (a) What is the frequency of Brian's string if beats of \(2.0 \mathrm{Hz}\) are heard when the two bow the strings together? (b) Does Brian need to tighten or loosen his A string to get in tune with Jennifer? Explain.

Short Answer

Expert verified
Answer: The frequency of Brian's violin string is 438.0 Hz, and he should tighten his A string to match Jennifer's frequency.

Step by step solution

01

Part (a): Finding the frequency of Brian's string

The beat frequency is the difference between the frequencies of the two strings, which can be expressed as: \(|f_J - f_B| = f_{\text{beat}}\) We have: - Jennifer's frequency: \(f_J = 440.0 \mathrm{Hz}\) - Beat frequency: \(f_{\text{beat}} = 2.0 \mathrm{Hz}\) We set up two equations to consider each case when Brian's frequency is higher or lower: 1. \(f_B = f_J - f_{\text{beat}}\) 2. \(f_B = f_J + f_{\text{beat}}\) Now calculate both possibilities: 1. \(f_B^1 = 440.0 - 2.0 = 438.0 \mathrm{Hz}\) 2. \(f_B^2 = 440.0 + 2.0 = 442.0 \mathrm{Hz}\) Since we are told Brian's A string is tuned to a slightly lower frequency than Jennifer's, we select the first case as the correct one, so: \(f_B = 438.0 \mathrm{Hz}\)
02

Part (b): Determining whether to tighten or loosen the string

To match Jennifer's frequency, Brian needs to get his frequency closer to Jennifer's, which is 440.0 Hz. Currently, Brian's frequency is 438.0 Hz, so he needs to increase it to get in tune with Jennifer. To increase the frequency of a string, one must tighten it. Thus, Brian needs to tighten his A string.

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Most popular questions from this chapter

During a thunderstorm, you can easily estimate your distance from a lightning strike. Count the number of seconds that elapse from when you see the flash of lightning to when you hear the thunder. The rule of thumb is that 5 s clapse for each mile of distance. Verify that this rule of thumb is (approximately) correct. (One mile is \(1.6 \mathrm{km}\) and light travels at a speed of $3 \times 10^{8} \mathrm{m} / \mathrm{s} .$ )
An intensity level change of \(+1.00 \mathrm{dB}\) corresponds to what percentage change in intensity?
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In this problem, you will estimate the smallest kinetic energy of vibration that the human ear can detect. Suppose that a harmonic sound wave at the threshold of hearing $\left(I=1.0 \times 10^{-12} \mathrm{W} / \mathrm{m}^{2}\right)\( is incident on the eardrum. The speed of sound is \)340 \mathrm{m} / \mathrm{s}\( and the density of air is \)1.3 \mathrm{kg} / \mathrm{m}^{3} .$ (a) What is the maximum speed of an element of air in the sound wave? [Hint: See Eq. \((10-21) .]\) (b) Assume the eardrum vibrates with displacement \(s_{0}\) at angular frequency \(\omega ;\) its maximum speed is then equal to the maximum speed of an air element. The mass of the eardrum is approximately \(0.1 \mathrm{g} .\) What is the average kinetic energy of the eardrum? (c) The average kinetic energy of the eardrum due to collisions with air molecules in the absence of a sound wave is about \(10^{-20} \mathrm{J}\) Compare your answer with (b) and discuss.
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