Question

A piano tuner sounds two strings simultancously. One has been previously tuned to vibrate at \(293.0 \mathrm{Hz}\). The tuner hears 3.0 beats per second. The tuner increases the tension on the as-yet untuned string, and now when they are played together the beat frequency is \(1.0 \mathrm{s}^{-1} .\) (a) What was the original frequency of the untuned string? (b) By what percentage did the tuner increase the tension on that string?

Short answer

Answer: The original frequency of the untuned string is 296 Hz and the percentage increase in tension on the untuned string is approximately 0.036%.

Step-by-step solution

Find the original frequency of the untuned string.

Given that the beat frequency with the untuned string is 3.0 Hz, we can write this as the absolute difference in the frequencies of the two strings: \(|f_1 - f_2| = 3\) Hz Where \(f_1\) is the frequency of the tuned string (293.0 Hz), and \(f_2\) is the unknown frequency of the untuned string. Since the untuned string frequency decreases when the tension is increased, we can deduce that \(f_2 > f_1\). Then we can rewrite the equation as: \(f_2 = f_1 + 3\) Hz Now substitute the given frequency of the tuned string into the equation: \(f_2 = 293.0 + 3 = 296 \mathrm{Hz}\) So, the original frequency of the untuned string is 296 Hz.

Calculate the new frequency of the untuned string.

We're given that when the tension is increased, the beat frequency reduces to 1.0 Hz. We can use the beat frequency formula again to find the new frequency of the untuned string: \(|f_1 - f_3| = 1\) Hz Where \(f_1\) is the frequency of the tuned string (293.0 Hz) and \(f_3\) is the unknown new frequency of the untuned string. Now we know that \(f_3 < f_1\). Then we can rewrite the equation as: \(f_3 = f_1 - 1\) Hz Substitute the given frequency of the tuned string into the equation: \(f_3 = 293.0 - 1 = 292 \mathrm{Hz}\) So, the new frequency of the untuned string after increasing the tension is 292 Hz.

Find the percentage increase in tension on the untuned string.

Now, we will use the frequency-tension relation to find the percentage increase in tension. The formula is: \(f = \sqrt{\frac{T}{\mu}}\) Where \(f\) is the frequency, \(T\) is the tension, and \(\mu\) is the linear density of the string. Since the linear density of the string doesn't change, we can write the equation for original untuned and tuned untuned strings as: \(f_2 = \sqrt{\frac{T_2}{\mu}}\) and \(f_3 = \sqrt{\frac{T_3}{\mu}}\) Now we can express the relation between percentage increase in tension and the frequencies: \(\frac{T_3 - T_2}{T_2} = \frac{(\sqrt{\frac{T_3}{\mu}})^2 - (\sqrt{\frac{T_2}{\mu}})^2}{(\sqrt{\frac{T_2}{\mu}})^2}\) By plugging in the values of \(f_2\) and \(f_3\) obtained in Steps 1 and 2: \(\frac{T_3 - T_2}{T_2} = \frac{(292)^2 - (296)^2}{(296)^2}\) Calculate the difference: \(\frac{T_3 - T_2}{T_2} = \frac{-32}{296^2}\) Now, multiply both sides by 100 to get the percentage increase in tension: \(Percent \ Increase = \frac{-32}{296^2} \times 100\) Calculate the value: \(Percent \ Increase \approx -0.036 \%\) The tuner increased the tension on the untuned string by approximately 0.036%.