Question

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

• A. $536\text{Hz}$
• B. $537\text{Hz}$
• C. $523\text{Hz}$
• D. $524\text{Hz}$

Answer 1

The correct option is D $524\text{Hz}$

Given:

$f_A=530\text{Hz};f_B=?$

As we know that beat frequency defined as

$\text{Beat frequency}=|f_A-f_B|$

So, given that

$|f_A-f_B|=6\text{Hz}$ and $f_A=530\text{Hz}$

$\therefore f_B=536\text{Hz}or524\text{Hz}$

Since, frequency is directly proportional to square root of tension $T$ in string.

$f\propto \sqrt{T}$

So when the tension in $\text{B}$ is slightly decrease, then frequency of $\text{B}$ also decreases. And given that beat frequency increases to $7\text{Hz}$.

It is only possible when $f_B=524\text{Hz},$ because for this value beat frequency will increase when $f_B$ will decrease.

So, $f_B=524\text{Hz}$

Hence, option $\left(d\right)$ is correct.