Question

A violin is tuned by adjusting the tension in the strings. Brian's A string is tuned to a slightly lower frequency than Jennifer's, which is correctly tuncd to \(440.0 \mathrm{Hz}\) (a) What is the frequency of Brian's string if beats of \(2.0 \mathrm{Hz}\) are heard when the two bow the strings together? (b) Does Brian need to tighten or loosen his A string to get in tune with Jennifer? Explain.

Short answer

Answer: The frequency of Brian's violin string is 438.0 Hz, and he should tighten his A string to match Jennifer's frequency.

Step-by-step solution

Part (a): Finding the frequency of Brian's string

The beat frequency is the difference between the frequencies of the two strings, which can be expressed as: \(|f_J - f_B| = f_{\text{beat}}\) We have: - Jennifer's frequency: \(f_J = 440.0 \mathrm{Hz}\) - Beat frequency: \(f_{\text{beat}} = 2.0 \mathrm{Hz}\) We set up two equations to consider each case when Brian's frequency is higher or lower: 1. \(f_B = f_J - f_{\text{beat}}\) 2. \(f_B = f_J + f_{\text{beat}}\) Now calculate both possibilities: 1. \(f_B^1 = 440.0 - 2.0 = 438.0 \mathrm{Hz}\) 2. \(f_B^2 = 440.0 + 2.0 = 442.0 \mathrm{Hz}\) Since we are told Brian's A string is tuned to a slightly lower frequency than Jennifer's, we select the first case as the correct one, so: \(f_B = 438.0 \mathrm{Hz}\)

Part (b): Determining whether to tighten or loosen the string

To match Jennifer's frequency, Brian needs to get his frequency closer to Jennifer's, which is 440.0 Hz. Currently, Brian's frequency is 438.0 Hz, so he needs to increase it to get in tune with Jennifer. To increase the frequency of a string, one must tighten it. Thus, Brian needs to tighten his A string.