Question

The middle-C key (key 40 ) on a piano corresponds to a fundamental frequency of about \(262 \mathrm{~Hz}\), and the soprano-C 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

Answer: The ratio of the tensions in the strings of the middle-C key (key 40) and soprano-C key (key 64) is 1:16.

Step-by-step solution

Write the formula for the fundamental frequency of a vibrating string

The formula for the fundamental frequency of a vibrating string is given by: $$ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} $$ where: - \(f\) is the fundamental frequency, - \(L\) is the length of the string, - \(T\) is the tension in the string, and - \(\mu\) is the linear density of the string.

Set up a ratio of the frequencies

We are given the fundamental frequencies for keys 40 and 64 as \(f_{40} = 262 \mathrm{~Hz}\) and \(f_{64} = 1046.5 \mathrm{~Hz}\), respectively. We can set up a ratio of their frequencies, as follows: $$ \frac{f_{40}}{f_{64}} = \frac{262}{1046.5} $$

Set up equations for each key's tension

We can rewrite the fundamental frequency formula for both keys: $$ f_{40} = \frac{1}{2L} \sqrt{\frac{T_{40}}{\mu}} \qquad \text{and} \qquad f_{64} = \frac{1}{2L} \sqrt{\frac{T_{64}}{\mu}} $$ Since the strings are identical in density and length, we know that \(L\) and \(\mu\) are the same for both strings.

Solve for the tension ratio

We can divide the equations from step 3 to eliminate \(L\) and \(\mu\): $$ \frac{f_{40}}{f_{64}} = \frac{\frac{1}{2L} \sqrt{\frac{T_{40}}{\mu}}}{\frac{1}{2L} \sqrt{\frac{T_{64}}{\mu}}} $$ This simplifies to: $$ \frac{f_{40}}{f_{64}} = \sqrt{\frac{T_{40}}{T_{64}}} $$ Now, square both sides of the equation: $$ \left(\frac{f_{40}}{f_{64}}\right)^2 = \frac{T_{40}}{T_{64}} $$ Plug in the frequency ratio obtained in step 2: $$ \left(\frac{262}{1046.5}\right)^2 = \frac{T_{40}}{T_{64}} $$

Calculate the result

Evaluate the left side of the equation to obtain the ratio of the tensions: $$ \frac{T_{40}}{T_{64}} = 0.0625 $$ So, the ratio of the tensions in the two strings is \(0.0625:1\) or \(1:16\), meaning the tension in key 64 is 16 times greater than the tension in key 40.