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A wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $45 \mathrm{~Hz}$. The mass of the wire is $3.5 \times 10^{-2} \mathrm{~kg}$ and its linear mass density is $4 \times 10^{-2} \mathrm{~kg} / \mathrm{m}$. What is (a) the speed of the transverse wave on the string and (b) the tension in the string?
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Verified Answer
(a) Given, $v=45 \mathrm{~Hz}, M=3.5 \times 10^{-2} \mathrm{~kg}$,
$$
\begin{aligned}
&m=4 \times 10^{-2} \mathrm{~kg} / \mathrm{m} \\
&\therefore l=\frac{M}{m}=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}}=\frac{7}{8} \mathrm{~m}
\end{aligned}
$$
As it is vibrating in fundamental mode
$$
\frac{\lambda}{2}=l=\frac{7}{8} \quad \therefore \quad \lambda=\frac{7}{4}=1.75 \mathrm{~m} .
$$
Speed $=v=v \lambda=45 \times 1.75=78.75 \mathrm{~m} / \mathrm{s}$
(b)
$$
\begin{aligned}
v &=\sqrt{\frac{T}{m}} \Rightarrow T=v^2 \times m \\
&=(78.5)^2 \times 4 \times 10^{-2}=248.06 \mathrm{~N}
\end{aligned}
$$
$$
\begin{aligned}
&m=4 \times 10^{-2} \mathrm{~kg} / \mathrm{m} \\
&\therefore l=\frac{M}{m}=\frac{3.5 \times 10^{-2}}{4 \times 10^{-2}}=\frac{7}{8} \mathrm{~m}
\end{aligned}
$$
As it is vibrating in fundamental mode
$$
\frac{\lambda}{2}=l=\frac{7}{8} \quad \therefore \quad \lambda=\frac{7}{4}=1.75 \mathrm{~m} .
$$
Speed $=v=v \lambda=45 \times 1.75=78.75 \mathrm{~m} / \mathrm{s}$
(b)
$$
\begin{aligned}
v &=\sqrt{\frac{T}{m}} \Rightarrow T=v^2 \times m \\
&=(78.5)^2 \times 4 \times 10^{-2}=248.06 \mathrm{~N}
\end{aligned}
$$
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