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A bead of mass $\mathrm{m}$ is attached to the mid-point of a taut, weightless string of length $\ell$ and placed on a frictionless horizontal table.
Under a small trverse displacement $x$, as shown, if the tension in the string is $T$, then the frequency of oscillation is-
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Under a small trverse displacement $x$, as shown, if the tension in the string is $T$, then the frequency of oscillation is-
Solution:
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Verified Answer
The correct answer is:
$\frac{1}{2 \pi} \sqrt{\frac{4 \mathrm{~T}}{\mathrm{~m} \ell}}$
Let the angle of $\mathrm{T}$ with the vertical be $\theta$ then F.B.D.
$\therefore 2 \mathrm{~T} \cos \theta=\mathrm{ma}$
also $\cos \theta=\frac{x}{\sqrt{x^{2}+\left(\frac{\ell}{2}\right)^{2}}}$
given $\ell>>x \quad \therefore \quad \frac{\ell^{2}}{4}+x^{2} \simeq \frac{\ell^{2}}{4}$
$\therefore \mathrm{a}=\frac{-2 \mathrm{Tx}}{\mathrm{m}\left(\frac{\ell}{2}\right)}$ (negative sign for restoring force)
or $\mathrm{a}=-\left(\frac{4 \mathrm{~T}}{\mathrm{~m} \ell}\right) \mathrm{x}$ also this is similar to the equation of $\mathrm{SHM}$ i.e. $\mathrm{a}=-\omega^{2} \mathrm{x}$
$$
\begin{array}{l}
\therefore \omega=\sqrt{\frac{4 T}{m \ell}} \& f=\frac{\omega}{2 \pi} \\
\therefore f=\frac{1}{2 \pi} \sqrt{\frac{4 T}{m \ell}}
\end{array}
$$
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