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A sitar wire is replaced by another wire of same length and material but of three times the earlier radius. If the tension in the wire remains the same, by what factor will the frequency change?
Solution:
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Verified Answer
The wire is stretched both end so frequency of stretched wire is
$$
v=\frac{n}{2 l} \sqrt{\frac{T}{m}}
$$
Mass of wire, $M=v p=A l \rho=\pi r^2 l \rho$
Mass per unit length
$$
m=\frac{\text { Mass of wire }}{\text { Length }}=\frac{\left(\pi \pi^2 l\right) \rho}{l}=\pi r^2 \rho
$$
So the ratio of $m_2$ and $m_1$ is
$$
\begin{aligned}
&\therefore \frac{m_2}{m_1}=\frac{\pi r_2^2 \rho}{\pi r_1^2 \rho} \\
&\frac{m_2}{m_1}=\left(\frac{r_2}{r_1}\right)^2 \Rightarrow \frac{r_2}{r_1}=\sqrt{\frac{m_2}{m_1}}
\end{aligned}
$$
We also know that
$$
v \propto \frac{1}{\sqrt{m}} \Rightarrow \frac{v_1}{v_2}=\frac{\sqrt{m_2}}{\sqrt{m_1}}
$$
So, $\frac{v_1}{v_2}=\frac{r_2}{r_1} \Rightarrow \frac{v_1}{v_2}=\frac{3}{1} \Rightarrow v_2=\frac{v_1}{3}$
$$
v=\frac{n}{2 l} \sqrt{\frac{T}{m}}
$$
Mass of wire, $M=v p=A l \rho=\pi r^2 l \rho$
Mass per unit length
$$
m=\frac{\text { Mass of wire }}{\text { Length }}=\frac{\left(\pi \pi^2 l\right) \rho}{l}=\pi r^2 \rho
$$
So the ratio of $m_2$ and $m_1$ is
$$
\begin{aligned}
&\therefore \frac{m_2}{m_1}=\frac{\pi r_2^2 \rho}{\pi r_1^2 \rho} \\
&\frac{m_2}{m_1}=\left(\frac{r_2}{r_1}\right)^2 \Rightarrow \frac{r_2}{r_1}=\sqrt{\frac{m_2}{m_1}}
\end{aligned}
$$
We also know that
$$
v \propto \frac{1}{\sqrt{m}} \Rightarrow \frac{v_1}{v_2}=\frac{\sqrt{m_2}}{\sqrt{m_1}}
$$
So, $\frac{v_1}{v_2}=\frac{r_2}{r_1} \Rightarrow \frac{v_1}{v_2}=\frac{3}{1} \Rightarrow v_2=\frac{v_1}{3}$
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