Question

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

• A. $8.1\,m$
• B. $2.1\,m$
• C. $1.1\,m$
• D. $5.1\,m$

Answer 1

Answer: (B)

$2.1\,m$

Answer 2

$\frac{nv}{2\ell} = 420$
$\frac{\left(n+1\right)v}{2\ell} = 490$
$\frac{v}{2\ell} = 70$
$\ell = \frac{v}{140} = \frac{1}{140}\sqrt{\frac{540}{6\times10^{-3}}} = \frac{1}{140}\sqrt{90\times10^{3}}$
$\ell = \frac{300}{140} = 2.142$