Question

A guitar wire $A$ vibrates at a fundamental frequency $600\text{Hz}$. A second identical wire $B$ (having fundamental frequency less than that of $A$) produces $6$ beats per second with it when the tension in $B$ is slightly decreased. Find the ratio of the tension in $A$ to the tension in $B$.

• A. $0.98$
• B. $1.25$
• C. $1.02$
• D. $0.75$

Answer 1

The correct option is C $1.02$

Given,
$f_b=6\text{Hz}$
We know, beat frequency
$f_b=|f_1-f_2|$
$\Rightarrow f_2=f_1\pm f_b$
Frequency of wire $A$,
$f_1=600\text{Hz}$ ............(1)
$\Rightarrow f_2=600\pm 6\text{Hz}$
So, frequency of wire $B$ can be,
$f_2=606\text{Hz}$ or $594\text{Hz}$
We know that, frequency,
$f=\frac{n}{2l}\sqrt{\frac{T}{\mu }}$ ...........(2)
As the tension in the wire $B$ is slightly decreased, therefore its frequency will also decrease and will be less than the frequency of wire $A$.
Hence, the frequency of wire $B$ will be
$f_2=594\text{Hz}$ ...........(3)
For identical wire, from (2), ratio of fundamental frequency,
$\frac{f_1}{f_2}=\sqrt{\frac{T_1}{T_2}}$
$\Rightarrow \frac{T_1}{T_2}=\frac{f_1^2}{f_2^2}$
From (1) and (3)
$\Rightarrow \frac{T_1}{T_2}=\frac{{600}^2}{{594}^2}=1.02$