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 \(1:16\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following transverse waves has the greatest power? a) a wave with velocity \(v\), amplitude \(A\), and frequency \(f\) b) a wave of velocity \(v\), amplitude \(2 A\), and frequency \(f / 2\) c) a wave of velocity \(2 v\), amplitude \(A / 2\), and frequency \(f\) d) a wave of velocity \(2 v\), amplitude \(A\), and frequency \(f / 2\) e) a wave of velocity \(v\), amplitude \(A / 2\), and frequency \(2 f\)

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?

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?

Two steel wires are stretched under the same tension. The first wire has a diameter of \(0.500 \mathrm{~mm}\), and the second wire has a diameter of \(1.00 \mathrm{~mm}\). If the speed of waves traveling along the first wire is \(50.0 \mathrm{~m} / \mathrm{s}\), what is the speed of waves traveling along the second wire?

A string is made to oscillate, and a standing wave with three antinodes is created. If the tension in the string is increased by a factor of 4 a) the number of antinodes increases. b) the number of antinodes remains the same. c) the number of antinodes decreases. d) the number of antinodes will equal the number of nodes.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free