Answer to Question #110020 in Optics for Jason Crary

Question

Two guitar strings have the same known length L, and same known linear mass density mu, vibrate at their common fundamental frequency of f1 Hz. you would like to know the frequency, f1, of the note you hear, but you don't have perfect pitch. Undeterred, you decide to use physics to find it. Both strings originally have some unknown T. You can increase the tension of one of the strings such that you known its tension is exactly 10% greater than that for the other one. Use your knowledge of interference and beats to find the orignally frequency, f1, in terms of the number of beats per second that you hear, n

Answer 1

As per the given question,

The length of the guitar strings are same,$l_1=l_2=L$

Linear mass density of the both string is also same,

$\mu_1=\mu_2=\mu$

Common fundamental frequency $f_1$ , let initial tension in the string of guitar was T

Initially,

$f_1=\dfrac{1}{2L}\sqrt{\dfrac{T}{\mu}}$

Now tension in the one string is getting increase by 10%,

New tension in the string is $T+\dfrac{10T}{100}=1.1T$

$f_2=\dfrac{1}{2L}\sqrt{\dfrac{1.1T}{\mu}}$

$f_2=\sqrt{1.1}f_1$

As per question, it is given that beat frequency $f_2-f_1=n--------(i)$

Beat frequency = $f_2-f_1=\sqrt{1.1}f_1-f_1=(\sqrt{1.1}-1)f_1=(1.0448-1)f_1=0.0448f_1$

=$0.0448f_1---------(ii)$

from the equation (i) and (ii),

$\Rightarrow n = 0.0448f_1$

$\Rightarrow f_1=\dfrac{n}{0.448}$

$\Rightarrow f_1=\dfrac{1000n}{448}=\dfrac{125n}{56}$

Hence fundamental frequency will be $(f_1)=\dfrac{125n}{56}Hz$