Question

Two strings on a musical instrument are tuned to play at 392 Hz (G) and 494 Hz (B). (a) What are the frequencies of the first two overtones for each string? (b) If the two strings have the same length and are under the same tension, what must be the ratio of their masses (c) If the strings, instead, have the same mass per unit length and are under the same tension, what is the ratio of their lengths (d) If their masses and lengths are the same, what must be the ratio of the tensions in the two strings?

Short answer

a) The first two overtone frequencies of the G string are, \(784\;{\rm{Hz}}\) , and \(1176\;{\rm{Hz}}\)respectively. Similarly, those of the B string are, r\(988\;{\rm{Hz}}\) , and \(1482\;{\rm{Hz}}\)respectively.

b) The ratio of the masses of the strings is 1.59.

c) The ratio of the lengths of the strings is 1.26.

d) The ratio of the tensions in the strings is 0.63.

Step-by-step solution

Given data in the question

The frequency of the G string is \({f_G} = \;392\;{\rm{Hz}}\).

The frequency of the B string is \({f_B} = \;494\;{\rm{Hz}}\).

Concepts used in the solution.

The frequency of the\({n^{th}}\)overtone is given by,

\({f_n} = \;\left( {n + 1} \right)\;f\)

The relationship between the frequency and mass of the string is given by,

\(f\; \propto \;\frac{1}{{\sqrt m }}\)

The relationship between the frequency and the length of the string is given by,

\(f \propto \;\frac{1}{L}\)

The relationship between the frequency and the tension in the string is given by,

\(f \propto \;\sqrt T \)

Calculation to obtain the desired result 

Part (a)

The frequency of the first overtone of the strings are,

\(\begin{aligned}{{f_{G1}} &= \;\left( {1 + 1} \right)\;{f_G}\\ &= \;2 \times \;392\;{\rm{Hz}}\\ &= \;784\;{\rm{Hz}}\\\\{f_{B1}} &= \;\left( {1 + 1} \right)\;{f_B}\\ &= \;2 \times \;494\;{\rm{Hz}}\\ &= \;988\;{\rm{Hz}}\end{aligned}\)

Therefore, the frequency of the first overtone of the G string is \(784\;{\rm{Hz}}\), and that of the B string is \(988\;{\rm{Hz}}\).

The frequency of the second overtone of the strings are,

\(\begin{aligned}{f_{G2}} &= \;\left( {2 + 1} \right)\;{f_G}\\ &= \;3 \times \;392\;{\rm{Hz}}\\ &= \;1176\;{\rm{Hz}}\\\\{f_{B2}} &= \;\left( {2 + 1} \right)\;{f_B}\\ &= \;3 \times \;494\;{\rm{Hz}}\\ &= \;1482\;{\rm{Hz}}\end{aligned}\)

Therefore, the frequency of the second overtone of the G string is \(1176\;{\rm{Hz}}\), and that of the B string is \(1482\;{\rm{Hz}}\).

Part (b)

From the relationship between the frequency and mass of the string, we get

\(\begin{aligned}\frac{{{m_G}}}{{{m_B}}} &= \;\frac{{f_B^2}}{{f_B^2}}\\ &= \;\frac{{{{\left( {494\;{\rm{Hz}}} \right)}^2}}}{{{{\left( {392\;{\rm{Hz}}} \right)}^2}}}\\ &= \;1.59\end{aligned}\)

Therefore, the ratio of their masses is 1.59.

Part (c)

From the relationship between the frequency and length of the string, we get

\(\begin{aligned}\frac{{{L_G}}}{{{L_B}}}\; &= \;\frac{{{f_B}}}{{{f_G}}}\\ &= \;\frac{{494\;{\rm{Hz}}}}{{392\;{\rm{Hz}}}}\\ &= \;1.26\end{aligned}\)

Therefore, the ratio of the lengths is 1.26.

Part (d)

From the relationship between the frequency and the tension in the string, we get

\(\begin{aligned}\frac{{{T_G}}}{{{T_B}}} &= \;{\left( {\frac{{{f_G}}}{{{f_B}}}} \right)^2}\\ &= \;{\left( {\frac{{392\;{\rm{Hz}}}}{{494\;{\rm{Hz}}}}} \right)^2}\\ &= \;0.63\end{aligned}\)

Therefore, the ratio of the tension is 0.63.