A bead of mass m is attached to the mid-point | vedclass

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(B) Resolving the tension $T$ in the string,the restoring force acting on the bead is $F = -2 T \sin 	heta$.From the geometry of the figure,$\sin 	h

Vedclass · Accepted Answer

$\frac{1}{2 \pi} \sqrt{\frac{4 T}{m l}}$

Vedclass · Answer

(B) Resolving the tension $T$ in the string,the restoring force acting on the bead is $F = -2 T \sin 	heta$.From the geometry of the figure,$\sin 	heta = \frac{x}{\sqrt{x^2 + (l/2)^2}}$.For a small displacement $x$,where $x \ll l/2$,we can approximate $\sin 	heta \approx 	an 	heta = \frac{x}{l/2} = \frac{2x}{l}$.Thus,the restoring force is $F = -2 T \left( \frac{2x}{l} ight) = -\left( \frac{4T}{ml} ight) m x$.Comparing this with the standard equation for simple harmonic motion,$F = -m \omega^2 x$,we get $\omega^2 = \frac{4T}{ml}$.The angular frequency is $\omega = \sqrt{\frac{4T}{ml}}$.The frequency of oscillation is $f = \frac{\omega}{2 \pi} = \frac{1}{2 \pi} \sqrt{\frac{4T}{ml}}$.

$A$ bead of mass $m$ is attached to the mid-point of a taut,weightless string of length $l$ and placed on a frictionless horizontal table. Under a small transverse displacement $x$,as shown in the figure,if the tension in the string is $T$,then the frequency of oscillation is:

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