Question:
A wire of length $\mathrm{L}$ and mass per unit length $6.0 \times 10^{-3}$ $\mathrm{kgm}^{-1}$ is put under tension of $540 \mathrm{~N}$. Two consecutive frequencies that it resonates at are: $420 \mathrm{~Hz}$ and $490 \mathrm{~Hz}$. Then $\mathrm{L}$ in meters is:
Correct Option: 1
Solution:
(1) Fundamental frequency, $f=70 \mathrm{~Hz}$.
The fundamental frequency of wire vibrating under tension $T$ is given by
$f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$
Here, $\mu=$ mass per unit length of the wire
$L=$ length of wire
$70=\frac{1}{2 L} \sqrt{\frac{540}{6 \times 10^{-3}}}$
$\Rightarrow \quad L \approx 2.14 \mathrm{~m}$
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