Question

Two guitar strings of 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

Here it is given that two guitar having same length

$l_1=l_2=l$ and it has also the same linear mass density

so we can write as

$\mu_1=\mu_2=\mu$ , we know that the frequency of the guitar note

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

as per the question if tension is increased by 10% then new tension become

T₁= T+0.1T=1.1 T

Now frequency will become

$f_2=\frac{1}{2L}\times \sqrt{\frac{T_1}{\mu}}$

$f_2=\sqrt{1.1}f_1$

Beat frequncy

$n=f_2-f_1=\sqrt{1.1}f_1-f_1=1.0488 f_1$

$f_1=\frac {n}{1.0488}=\frac{125n}{56}$