Question

In a guitar, two strings A and B made of same material are slightly out of tune and produce beats of frequency $6Hz$. When tension in B is slightly decreased, the be at frequency increases to $7Hz$. If the frequency of A is $530Hz$, 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 $6HZ$
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 $f_B^{\prime }$
Now, difference of $f_A$ and $f_B^{\prime }=7Hz$ (increases)
So, $f_A=f_B$
$f_A-f_B=6Hz$
$f_A=530Hz$
$f_B=524Hz$ (original)