A guitarist tunes a string such that it is exactly in tune with a second string of equal linear mass density and length.Both vibrate with a fundamental frequency of 110.0Hz.He decreases the tension in one string so that he hears beats of 4.6Hz.What is the ratio of the tension of the first string to that of the second?
A)0.918
B)0.921
C)0.958
D)0.960
E)0.979

Question

Quizplus · Accepted Answer

The frequency of beats is given by the absolute difference between the frequencies of the two strings. When the tension in one string is decreased, the frequency of that string changes, leading to the beats phenomenon. The fundamental frequency of a string is given by $f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$, where $L$ is the length of the string, $T$ is the tension, and $\mu$ is the linear mass density. Since the lengths and linear mass densities of the strings are equal, the ratio of the frequencies is equal to the square root of the ratio of the tensions. Given that the beat frequency is 4.6 Hz, one of the strings now has a frequency of either $110 + 4.6 = 114.6$ Hz or $110 - 4.6 = 105.4$ Hz. Using the formula for the fundamental frequency and solving for the ratio of the tensions, we find that the ratio of the frequencies squared is equal to the ratio of the tensions. Let's assume the decreased tension results in the 105.4 Hz frequency. The ratio of the frequencies squared is $(\frac{105.4}{110})^2$, which gives the ratio of the tensions. Calculating this gives:$(\frac{105.4}{110})^2 = 0.918$Therefore, the ratio of the tension of the first string to that of the second is approximately 0.918.

A Guitarist Tunes a String Such That It Is Exactly