Question

(II) Two piano strings are supposed to be vibrating at 220 Hz, but a piano tuner hears three beats every 2.5 s when they are played together. (a) If one is vibrating at 220.0 Hz, what must be the frequency of the other (is there only one answer)? (b) By how much (in percent) must the tension be increased or decreased to bring them in tune?

Short answer

1. The beat frequency of the other string can be either \(218.8\;{\rm{Hz}}\) or\(221.2\;{\rm{Hz}}\).
2. The tension is increased by \(1.1\% \) at frequency \(218.8\;{\rm{Hz}}\) and decreased by \(1.1\% \) at frequency \(221.2\;{\rm{Hz}}\).

Step-by-step solution

Determination of the fundamental frequency

The string’s fundamental frequency can be obtained by the ratio of the whole square root of the string’s tension and linear density to twice the string’s length.

Given information

Given data:

The frequency of vibration is \(f' = 220\;{\rm{Hz}}\).

The time required to hear three beats is \(T = 2.5\;{\rm{s}}\).

Evaluation of the beat frequency of the other string

(a)

The beat period of the single beat can be calculated as:

\(\begin{aligned}{c}t = \frac{T}{N}\\t = \frac{{2.5\;{\rm{s}}}}{3}\\t = 0.833\;{\rm{s}}\end{aligned}\)

As the beat frequency is inverse of the time period, it can be calculated as,

\(\begin{aligned}{c}{f_B} = \frac{1}{t}\\{f_B} = \frac{1}{{0.833\;{\rm{s}}}}\\{f_B} = 1.2\;{\rm{Hz}}\end{aligned}\)

The beat frequency of the other string can be calculated as:

\(\begin{aligned}{c}{f_B} = \left| {f' \pm f} \right|\\1.2\;{\rm{Hz}} = \left| {220\;{\rm{Hz}} \pm f} \right|\\f = 218.8\;{\rm{Hz}}\;{\rm{or}}\;221.2\;{\rm{Hz}}\end{aligned}\)

Thus, the beat frequency of the other string can be either \(218.8\;{\rm{Hz}}\) or \(221.2\;{\rm{Hz}}\).

Evaluation of the change in tension

(b)

From the relation of the fundamental frequency of the string, you have:

\(f \propto \sqrt {{F_T}} \) … (1)

Similarly, for the new fundamental frequency, you have:

\(f' \propto \sqrt {{{F'}_T}} \) … (2)

After dividing the equation (2) by (1), you get:

\(\begin{aligned}{c}\frac{{f'}}{f} = \frac{{\sqrt {{{F'}_T}} }}{{\sqrt {{F_T}} }}\\{\left( {\frac{{220\;{\rm{Hz}}}}{{218.8\;{\rm{Hz}}}}} \right)^2} = \left( {\frac{{{{F'}_T}}}{{{F_T}}}} \right)\\{{F'}_T} = 1.011{F_T}\end{aligned}\)

Similarly,

\(\begin{aligned}{c}\frac{{f'}}{f} = \frac{{\sqrt {{{F'}_T}} }}{{\sqrt {{F_T}} }}\\{\left( {\frac{{220\;{\rm{Hz}}}}{{221.2\;{\rm{Hz}}}}} \right)^2} = \left( {\frac{{{{F'}_T}}}{{{F_T}}}} \right)\\{{F'}_T} = 0.9892{F_T}\end{aligned}\)

Thus, the tension is increased by \(1.1\% \) at frequency \(218.8\;{\rm{Hz}}\) and decreased by \(1.1\% \) at frequency \(221.2\;{\rm{Hz}}\).