Question

In a guitar, two strings $A$ and $B$ made up of the same material produce beats of frequency $4\text{Hz}$. When the tension in $B$ is slightly decreased, the beat frequency increases to $6\text{Hz}$. If the frequency of $A$ is $530\text{Hz}$ then the original frequency of $B$ will be

• A. $526\text{Hz}$
• B. $546\text{Hz}$
• C. $530\text{Hz}$
• D. $542\text{Hz}$

Answer 1

The correct option is A $526\text{Hz}$

Frequency of string $A$,
$f_1=530\text{Hz}$
Frequency of string $B=f_2$
Beat frequency,
$f_b=4\text{Hz}$
We know, beat frequency
$f_b=|f_1-f_2|$
$\therefore f_1-f_2=4$ or $f_2-f_1=4$
We know that, frequency,
$f=\frac{n}{2l}\sqrt{\frac{T}{\mu }}$
As the tension in the string $B$ is decreased, therefore its frequency will also decrease.
Hence for beat frequency to increase with frequency of B decreasing, $f_1-f_2=4$ holds true
Therefore, $f_2=f_1-4=530-4=526\text{Hz}$