Question

Two strings are adjusted to vibrate at exactly 202 Hz. Then the tension in one string isincreased slightly. Afterward, three beats per second are heardwhen the strings vibrate at the same time. What is the newfrequency of the string that was tightened?

Answer 1

Concept and reason

The concepts needed to solve this question are the number of beats formed per second by two frequencies and frequency of the vibrations of the vibrating string.

Initially, write the equation for frequency of a vibrations of a vibrating string. After that, identify the relation that how the frequency changed with the tension in the string. Now, write the equation for number of beats formed per second by two frequencies. Finally, solve for the frequency of the tightened string by using above two relations.

Fundamentals

The expression for the frequency of the vibrations of a vibrating string is,

$f = \frac{1}{{2L}}\sqrt {\frac{T}{\mu }}$

Here, L is the vibrating length of the string, T is the tension in the string, and $\mu$ is the linear mass density of the material of the string.

The expression for the number n of beats formed per second by two frequencies is,

$n = \left| {{f_1} - {f_2}} \right|$

Here, ${f_1}$ is the first frequency and ${f_2}$ the second frequency.

The expression for the frequency of the vibrations of a vibrating string is,

$f = \frac{1}{{2L}}\sqrt {\frac{T}{\mu }}$

From this equation frequency of a string is directly proportional to the tension in it.

$f \propto \sqrt T$

The expression for the number n of beats formed per second by two frequencies is,

$n = \left| {{f_1} - {f_2}} \right|$

The relation between the frequency and tension in a string is,

$f \propto \sqrt T$

From this relation the frequency of the tightened string increases. Hence, beats will form by two waves one is due to first string and the second one that is due to the wave on the tightened string.

Therefore, the value of ${f_2}$ (frequency of tightened string) is more than the frequency of the first string ${f_1}$ . So, the equation $n = \left| {{f_1} - {f_2}} \right|$ can be written as follows:

$n = {f_2} - {f_1}$

Rearrange the equation for ${f_2}$ .

${f_2} = n + {f_1}$

Substitute $3{\rm{ /s}}$ for n and 202 Hz for ${f_1}$ .

$\begin{array}{c}\\{f_2} = 3{\rm{ /s}} + 202{\rm{ Hz}}\\\\ = 3{\rm{ /s}}\left( {\frac{{1{\rm{ Hz}}}}{{1{\rm{ /s}}}}} \right) + 202{\rm{ Hz}}\\\\ = 2{\rm{05 Hz}}\\\end{array}$

Ans:

The frequency of the tightened is 205 Hz.