Question

The frequency of the fundamental note in a wire stretched under tension$T$ is $v.$ If the tension is increased to $25T$, then the frequency of the fundamental note will be

• A. $25f$
• B. $5f$
• C. $10f$
• D. $f$

Answer 1

The correct option is B

$5f$

Step 1: Given data

Frequency of the fundamental note, $v=f$

Initial tension in the wire $\left(T_1\right)=T$

Increased tension in the wire$\left(T_2\right)=25T$

Step 2: Formula used

We know that the following formula represents the relation between frequency and tension.

$v=\frac{1}{2l}\sqrt{\frac{T}{m}},$where $v=$frequency, $T=$tension, $l=$length, and $m=$mass of unit length of the string

Step 3: Compute the new frequency

According to the formula, the frequency of a string is directly proportional to the square root of its tension. So,

$\begin{array}{rcl}v& \propto & \sqrt{T}\\ \frac{v_1}{v_2}& =& \sqrt{\frac{T_1}{T_2}}\\ \frac{f}{v_2}& =& \sqrt{\frac{T}{25T}}\\ v_2& =& 5f\end{array}$

Thus, the frequency of the fundamental note will be $5f$.

Hence, option B is the correct answer.