Question

In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency $6\,Hz$. When tension in B is slightly decreased, the be at frequency increases to $7\,Hz$. If the frequency of A is $530\,Hz$, the original frequency of B will be :

• A. 523 Hz
• B. 524 Hz
• C. 536 H z
• D. 537 H z

Answer 1

Answer: (B)

It is given, the difference of $f_{A}$ and $f_{B}$ is $6\,HZ$
Guitar string i.e. string is fixed from both ends

Frequency $\propto \sqrt{\text{Tension}}$
If tension in B slightly decrease then frequency of B decreases
If B is 536 Hz, as the frequency decreases, beats with A also decreases
If B is 524 Hz, as the frequency decreases, beats with A increases
If tension decreases, $f_{B}$ decreases and becomes $f'_{B}$
Now, difference of $f_{A}$ and $f'_{B}=7\,Hz$ (increases)
So, $f_{A}=f_{B}$
$f_{A}-f_{B}=6\,Hz$
$f_{A}=530\,Hz$
$f_{B}=524\,Hz$ (original)