Question

A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45\text{Hz}$. The mass of the wire is $3.5\times {10}^{-2}\text{kg}$ and its linear density is $4.0\times {10}^{-2}{\text{kg m}}^{-1}$. The tension in the string is approximately -

• A. $300.47\text{N}$
• B. $425.25\text{N}$
• C. $247.27\text{N}$
• D. $500\text{N}$

Answer 1

The correct option is C $247.27\text{N}$

Mass of wire $\left(m\right)=3.5\times {10}^{-2}\text{kg}$
Linear density $\left(\mu \right)=4\times {10}^{-2}\text{kg/m}$
$\mu =\frac{m}{L}$
$\Rightarrow L=\frac{m}{\mu }=\frac{3.5\times {10}^{-2}}{4\times {10}^{-2}}=\frac{7}{8}m$

In fundamental mode , $L=\frac{\lambda }{2}$
$\therefore \lambda =2L=\frac{7}{4}m$
Now velocity of transverse wave is,
$v=f\lambda =45\times \frac{7}{4}=78.75\text{m/s}$
Using $\Rightarrow v=\sqrt{\frac{T}{\mu }}$
$\Rightarrow T=\mu v^2$
$\therefore T=4\times {10}^{-2}\times (78.75{)}^2=248\text{N}$