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Chapter 15: Problem 55

The middle-C key (key 52 ) on a piano corresponds to a fundamental frequency of about \(262 \mathrm{~Hz},\) and the sopranoC key (key 64) corresponds to a fundamental frequency of \(1046.5 \mathrm{~Hz}\). If the strings used for both keys are identical in density and length, determine the ratio of the tensions in the two strings.

Short Answer

Expert verified
Answer: The ratio of tensions in the two strings is 1:16.

Step by step solution

01

Write the fundamental frequency formula for both keys

To find the ratio of the tensions of the two strings, first write the fundamental frequency formula for each of the keys (middle-C and soprano-C): Middle-C key: \(f_{1} = \frac{1}{2L} \sqrt{\frac{T_{1}}{\mu}}\) Soprano-C key: \(f_{2} = \frac{1}{2L} \sqrt{\frac{T_{2}}{\mu}}\)
02

Set up a proportion to compare the two strings

Since the two strings have the same density and length, the ratio of their fundamental frequencies should be proportional to the ratio of their tensions. We can write this proportion as: \(\frac{f_{1}}{f_{2}} = \frac{\frac{1}{2L} \sqrt{\frac{T_{1}}{\mu}}}{\frac{1}{2L} \sqrt{\frac{T_{2}}{\mu}}}\) Here, \(f_{1} = 262 \mathrm{~Hz}\) (middle-C frequency) and \(f_{2} = 1046.5 \mathrm{~Hz}\) (soprano-C frequency).
03

Simplify the proportion and solve for the tension ratio

Next, simplify the proportion: \(\frac{262}{1046.5} = \frac{\sqrt{\frac{T_{1}}{\mu}}}{\sqrt{\frac{T_{2}}{\mu}}}\) Square both sides of the equation to get rid of the square roots: \(\left(\frac{262}{1046.5}\right)^{2} = \frac{\frac{T_{1}}{\mu}}{\frac{T_{2}}{\mu}}\) Now, multiply both sides by \(T_{2}\) to isolate the ratio \(\frac{T_{1}}{T_{2}}\): \(\frac{T_{1}}{T_{2}} = \left(\frac{262}{1046.5}\right)^{2} \cdot 1\)
04

Calculate the tension ratio

Finally, calculate the tension ratio: \(\frac{T_{1}}{T_{2}} = \left(\frac{262}{1046.5}\right)^{2}\) \(\frac{T_{1}}{T_{2}} = \frac{1}{16}\) So, the ratio of the tensions in the two strings is \(1:16\).

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Most popular questions from this chapter

The largest tension that can be sustained by a stretched string of linear mass density \(\mu\), even in principle, is given by \(\tau=\mu c^{2},\) where \(c\) is the speed of light in vacuum. (This is an enormous value. The breaking tensions of all ordinary materials are about 12 orders of magnitude less than this.) a) What is the speed of a traveling wave on a string under such tension? b) If a \(1.000-\mathrm{m}\) -long guitar string, stretched between anchored ends, were made of this hypothetical material, what frequency would its first harmonic have? c) If that guitar string were plucked at its midpoint and given a displacement of \(2.00 \mathrm{~mm}\) there to produce the fundamental frequency, what would be the maximum speed attained by the midpoint of the string?

A small ball floats in the center of a circular pool that has a radius of \(5.00 \mathrm{~m}\). Three wave generators are placed at the edge of the pool, separated by \(120 .\). The first wave generator operates at a frequency of \(2.00 \mathrm{~Hz}\). The second wave generator operates at a frequency of \(3.00 \mathrm{~Hz}\). The third wave generator operates at a frequency of \(4.00 \mathrm{~Hz}\). If the speed of each water wave is \(5.00 \mathrm{~m} / \mathrm{s}\), and the amplitude of the waves is the same, sketch the height of the ball as a function of time from \(t=0\) to \(t=2.00 \mathrm{~s}\), assuming that the water surface is at zero height. Assume that all the wave generators impart a phase shift of zero. How would your answer change if one of the wave generators was moved to a different location at the edge of the pool?

A traveling wave propagating on a string is described by the following equation: $$ y(x, t)=(5.00 \mathrm{~mm}) \sin \left(\left(157.08 \mathrm{~m}^{-1}\right) x-\left(314.16 \mathrm{~s}^{-1}\right) t+0.7854\right) $$ a) Determine the minimum separation, \(\Delta x_{\min }\), between two points on the string that oscillate in perfect opposition of phases (move in opposite directions at all times). b) Determine the separation, \(\Delta x_{A B}\), between two points \(A\) and \(B\) on the string, if point \(B\) oscillates with a phase difference of 0.7854 rad compared to point \(A\). c) Find the number of crests of the wave that pass through point \(A\) in a time interval \(\Delta t=10.0 \mathrm{~s}\) and the number of troughs that pass through point \(B\) in the same interval. d) At what point along its trajectory should a linear driver connected to one end of the string at \(x=0\) start its oscillation to generate this sinusoidal traveling wave on the string?

Write the equation for a sinusoidal wave propagating in the negative \(x\) -direction with a speed of \(120 . \mathrm{m} / \mathrm{s}\), if a particle in the medium in which the wave is moving is observed to swing back and forth through a \(6.00-\mathrm{cm}\) range in \(4.00 \mathrm{~s}\). Assume that \(t=0\) is taken to be the instant when the particle is at \(y=0\) and that the particle moves in the positive \(y\) -direction immediately after \(t=0\).

One of the main things allowing humans to determine whether a sound is coming from the left or the right is the fact that the sound will reach one ear before the other. Given that the speed of sound in air is \(343 \mathrm{~m} / \mathrm{s}\) and that human ears are typically \(20.0 \mathrm{~cm}\) apart, what is the maximum time resolution for human hearing that allows sounds coming from the left to be distinguished from sounds coming from the right? Why is it impossible for a diver to be able to tell from which direction the sound of a motor boat is coming? The speed of sound in water is \(1.50 \cdot 10^{3} \mathrm{~m} / \mathrm{s}\).
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