Question

The length of the wire shown in figure between the pulley and fixed support is $1.5\text{m}$ and mass is $12.0\text{g}$. The frequency of vibration with which the wire vibrates in two loops leaving the middle point of the wire between the pulley and the fixed support, is: [Assume the pulley acts as a fixed support and consider $g=10{\text{m/s}}^2$]

• A. $10\text{Hz}$
• B. $30\text{Hz}$
• C. $100\text{Hz}$
• D. $70\text{Hz}$

Answer 1

The correct option is C $100\text{Hz}$

The string behaves as if fixed at both ends.
Standing wave with 2 loops can be considered as


$\Rightarrow \lambda =L$

We know that
Frequency, $f=\frac{\text{Velocity}}{\lambda }=\frac{v}{\lambda }$

From FBD of block:


Tension in the string is

$T=$ Tension $=18g\text{N}$

Mass per unit length of the string is

$\mu =\frac{12\times {10}^{-3}}{1.5}\text{kg/m}$

Velocity of wave in the string is

$v=\sqrt{\frac{T}{\mu }}$

$\Rightarrow v=\sqrt{\frac{18\times 10\times 1.5}{12\times {10}^{-3}}}=150\text{m/s}$

$\therefore f=\frac{v}{\lambda }=\frac{v}{L}=\frac{150}{1.5}=100\text{Hz}$