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  4. A wire of length L and mass per unit length 6.0 × 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 :

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Q. 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 :

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Solution:

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