m = mass per unit length of the string R = Radius of the loop $\omega=$ angular velocity, V = linear velocity of the string
Consider one half of the string as shown in figure. The half loop experiences cetrifugal force at every point, away from centre, which is balanced by tension 2T. Consider an element of angular part $\text{d}\theta$ at angle $\theta.$ Consider another element symmetric to this centrifugal force experienced by the element $=(\text{mRd}\theta)\omega^2\text{R}.$$\big($Length of element $=\text{Rd}\theta,\ \text{mass}=\text{mRd}\theta\big)$
Resolving into rectangular components net force on the two symmetric elements,$\text{DF}=2\text{mR}^2\text{d}\theta\omega^2\sin\theta$ [horizontal components cancels each other]
So, total $\text{F}=\int\limits_0^{\frac{\pi}{2}}2\text{mR}^2\omega^2\sin\theta\text{d}\theta=2\text{mR}^2\omega^2[-\cos]\Rightarrow2\text{mR}^2\omega^2$ Again, $2\text{T}=2\text{mR}^2\omega^2\ \Rightarrow\text{T}=\text{mR}^2\omega^2$ Velocity of transverse vibration $\text{v}=\sqrt{\frac{\text{T}}{\text{m}}}=\omega\text{R}=\text{v}$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
