Question

A wire of length \(1.2\,\text{m}\) is subjected to a tension of \(240\,\text{N}\). If the frequencies of two successive modes of vibration of the wire are \(200\,\text{Hz}\) and \(250\,\text{Hz}\), then the mass of the wire is

• A. \(10\,\text{g}\)
• B. \(20\,\text{g}\)
• C. \(30\,\text{g}\)
• D. \(40\,\text{g}\)

Hint:

For a stretched string, \[ \boxed{ f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu}}. } \] The difference between two successive frequencies is \[ \boxed{ f_{n+1}-f_n=\frac{1}{2L}\sqrt{\frac{T}{\mu}}. } \]

Answer 1

The Correct Option is B

Step 1: Use the formula for the frequency of a stretched string. For the \(n^{\text{th}}\) mode, \[ f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu}}, \] where \[ \mu=\frac{m}{L} \] is the mass per unit length. For successive modes, \[ f_{n+1}-f_n = \frac{1}{2L}\sqrt{\frac{T}{\mu}}. \]

Step 2: Determine the wave speed. Given, \[ f_{n+1}-f_n = 250-200 = 50\,\text{Hz}. \] Hence, \[ 50 = \frac{1}{2(1.2)} \sqrt{\frac{240}{\mu}}. \] Therefore, \[ \sqrt{\frac{240}{\mu}} = 50\times2.4 = 120. \] Squaring, \[ \frac{240}{\mu} = 120^2 = 14400. \] Thus, \[ \mu = \frac{240}{14400} = \frac1{60}\,\text{kg m}^{-1}. \]

Step 3: Calculate the mass of the wire. Since \[ m=\mu L, \] \[ m = \frac1{60}\times1.2 = 0.02\,\text{kg} = 20\,\text{g}. \] Hence, \[ \boxed{m=20\,\text{g}.} \] Therefore, the correct option is \(\boxed{(B)}\).