Question

A segment of wire vibrates with a fundamental frequency of $450Hz$ under a tension of$9kg–wt$. Then, tension at which the fundamental frequency of the same wire becomes $900Hz$is

• A. $36kg–wt$
• B. $27kg–wt$
• C. $18kg–wt$
• D. $72kg–wt$

Answer 1

The correct option is A

$36kg–wt$

Step 1. Given data

Frequency, $f_1=450Hz$

$f_2=900Hz$

Tension, $T_1=9kg–wt$

Step 2. Find the tension $T_2$:

As we know, the fundamental frequency of the standing wave is,

$f=\frac{1}{2L}\sqrt{\frac{T}{m}}$

Where,$T=$tension in the string

$m=$linear density of the string

$f=$fundamental frequency

$L=$length of the string

$\Rightarrow$ $\begin{array}{rcl}f& \propto & \sqrt{T}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{f_1}{f_2}& =& \sqrt{\frac{T_1}{T_2}}\end{array}$

$\Rightarrow$$\begin{array}{rcl}\frac{450}{900}& =& \sqrt{\frac{9}{T_2}}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{1}{4}& =& \frac{9}{T_2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{T}_2& =& 36kg–wt\end{array}$

Hence, (A) option is correct.