Question

The fundamental frequency of a sonometer wire increases by \(4\) Hz if the tension in the string is increased by \(21\%\) keeping the length of the wire constant. What will be the new fundamental frequency of the wire if its length is increased by \(25\%\) while keeping the original tension in the wire?

• A. Will remain the same
• B. \(32\) Hz
• C. \(34\) Hz
• D. \(35\) Hz

Hint:

For a sonometer wire: \[ f\propto \sqrt{T} \] and \[ f\propto \frac1L. \] A \(21\%\) increase in tension produces a \(10\%\) increase in frequency because \[ \sqrt{1.21}=1.1. \]

Answer 1

The Correct Option is B

Concept: The fundamental frequency of a stretched string is \[ f=\frac{1}{2L}\sqrt{\frac{T}{\mu}} \] Thus, \[ f\propto \sqrt{T} \] when \(L\) is constant, and \[ f\propto \frac1L \] when \(T\) is constant.

Step 1: Determine the original frequency. When tension is increased by \(21\%\), \[ T'=1.21T. \] Therefore, \[ f' = f\sqrt{1.21} = 1.1f. \] Given that the increase in frequency is \(4\) Hz, \[ 1.1f-f=4 \] \[ 0.1f=4 \] \[ f=40\ \text{Hz}. \]

Step 2: Increase the length by \(25\%\). \[ L'=1.25L. \] Keeping the original tension unchanged, \[ f_{\text{new}} = \frac{f}{1.25} \] \[ = \frac{40}{1.25} \] \[ = 32\ \text{Hz}. \] \[\begin{aligned} \boxed{32\ \text{Hz}} \end{aligned}\] Hence, option \(\mathbf{(B)}\) is correct.