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A uniform string is vibrating with a fundamental frequency ' $f$ '. The new frequency, if radius \& length both are doubled would be
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Verified Answer
The correct answer is:
$\frac{f}{4}$
Fundamental frequency of a vibrating string is expressed as
$$
f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}=\frac{1}{L D} \sqrt{\frac{T}{\pi \rho}}
$$
where,
$$
\begin{aligned}
& D= \text { diameter of string } \\
& \rho= \text { density of the } \\
& \text { material of string }
\end{aligned}
$$
As length $L$ and radius are doubled, the new frequency
$$
f^{\prime}=\frac{1}{(2 L)(2 D)} \sqrt{\frac{T}{\pi \rho}}=\frac{1}{4} f
$$
$$
f=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}=\frac{1}{L D} \sqrt{\frac{T}{\pi \rho}}
$$
where,
$$
\begin{aligned}
& D= \text { diameter of string } \\
& \rho= \text { density of the } \\
& \text { material of string }
\end{aligned}
$$
As length $L$ and radius are doubled, the new frequency
$$
f^{\prime}=\frac{1}{(2 L)(2 D)} \sqrt{\frac{T}{\pi \rho}}=\frac{1}{4} f
$$
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