Question

A wire of length $L$ and mass per unit length $6.0\times {10}^{-3}{\text{kgm}}^{-1}$ is put under tension of $540\text{N}$. Two consecutive frequencies that resonates with it are : $420\text{Hz}$ and $490\text{Hz}$. Then the value of $L$ (in meters) is:

• A. $\text{2.1 m}$
• B. $\text{1.1 m}$
• C. $\text{8.1 m}$
• D. $\text{5.1 m}$

Answer 1

The correct option is A $\text{2.1 m}$

Let the frequency of $p^{\text{th}}$ harmonic is $420\text{Hz}$ and that of $(p+1{)}^{\text{th}}$ harmonic is $490\text{Hz}$.

Then, $p\cdot f=420(f\to \text{fundamental frequency})$

And, $(p+1)\cdot f=490$

$\Rightarrow (p+1)\cdot f-p\cdot f=490-420$

$\Rightarrow f=70\text{Hz}$

The fundamental frequency of wire vibrating under tension $T$ is given by,

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

Here, $\mu =$ mass per unit length of wire
$L=$ length of wire

$70=\frac{1}{2L}\sqrt{\frac{540}{6\times {10}^{-3}}}$

$\Rightarrow L\approx 2.1\text{m}$

Hence, $\left(A\right)$ is the correct answer.