Question

The fundamental frequency of a segment of wire vibrating is $450Hz$ and it is under a tension of $9kg–wt$. Then, the 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 of a segment of wire vibrating $\left(f_1\right)=450Hz$

Increased frequency of wire$\left(f_2\right)=900Hz$

Tension of wire$=9kg–wt$

Step 2: Formula used:

Consider the following formula that represents the relationship between fundamental frequency and tension.

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

Where $f=$frequency, $L=$length, $T=$tension, and $\mu =$linear density

Step 3: Compute the tension:

From the formula, it is clear that the frequency$\left(f\right)$ of a string is directly proportional to the square root of its tension$\left(T\right)$. So,

$\begin{array}{rcl}f& \propto & \sqrt{T}\\ \frac{f_1}{f_2}& =& \sqrt{\frac{T_1}{T_2}}\\ \frac{450}{900}& =& \sqrt{\frac{9}{T_2}}\\ T_2& =& 4\times 9\\ T_2& =& 36kg-wt\end{array}$

So, the tension at which the fundamental frequency of the same wire becomes$900Hz$ is $36kg-wt$.

Hence, option A is the correct answer.