Question

Two identical guitar strings of length $L$ are each tuned exactly to $400\text{Hz}$. The tension in one of the strings is then increased by $2\%$. If they are now struck, what is the beat frequency between the fundamental modes of the two strings?

• A. $2\text{Hz}$
• B. $4\text{Hz}$
• C. $6\text{Hz}$
• D. $8\text{Hz}$

Answer 1

The correct option is B $4\text{Hz}$

Let $f_1$ be the fundamental frequency of first string which have higher tension and $f_2$ be the fundamental freqency of other string.
From formula
$f_1=\frac{1}{2L}\sqrt{\frac{T_1}{{\mu }_1}}$
and
$f_2=\frac{1}{2L}\sqrt{\frac{T_2}{{\mu }_2}}$
Since, both strings are identical.
So, ${\mu }_1={\mu }_2$
$\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}$
Let, $T_2=T$
As tension is increased by two percent,
$T_1=T+\frac{2}{100}T=1.02T$
So equation (i) becomes,
$\frac{f_1}{f_2}=\sqrt{\frac{1.02T}{T}}=\sqrt{1.02}=1.0099\approx 1.01$
$f_1=f_2\times 1.01$
$\Rightarrow f_1=400\times 1.01=404\text{Hz}[\because f_2=400\text{Hz}(\text{Given}\left)\right]$

Beat frequency $=|f_1-f_2|=404-400=4\text{Hz}$