Single Choice

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$$1.1\ m$$
B$$5.1\ m$$
C$$2.1\ m$$
Correct Answer
D$$8.1\ m$$

Solution

Given, $$f_1=420\,H_z$$ $$f_2=490\,H_z$$ Fundamental frequency $$=f_2-f_1=490-420=70\,Hz$$ We know, $$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{u}}$$ $$\Rightarrow l=\dfrac{1}{2l}\sqrt{\dfrac{540}{6\times 10^{-3}}}$$ $$l=\dfrac{1}{2\times 70}\sqrt{\dfrac{540}{6\times 10^{-3}}}=\dfrac{1}{2\times 70}\times \sqrt{90\times 10^3}=\dfrac{300}{140}$$ $$\Rightarrow \boxed{l\approx 2.14m}$$

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