Two masses m and m2 are connected at the two ends of a massless rigid rod of length l. The rod is suspended by a thin wire of torsional constant k at the centre of mass of the rod-mass system (see figure). Because of torsional constant k, the restoring torque is τ=kθ for angular displacement θ . If the rod is rotated by θ0 and released, the tension in it when it passes through its mean position will be: - Sri Chaitanya Infinity Learn Best Online Courses for NCERT Solutions, CBSE, ICSE, JEE, NEET, Olympiad and Class 6 to 12

From the figure,

$ω = \sqrt{\frac{k}{I}} = \sqrt{\frac{3 k}{m l^{2}}}$

$= Angular frequency \times Angular amplitude = {ωθ}_{0} = velocity at mean position$

$T = {mω}_{\max}^{2} r_{1} = {mω}_{\max}^{2} \frac{l}{3}$

$= {mω}^{2} {θ_{0}}^{2} \frac{l}{3}$

$= m \frac{3 k}{m l^{2}} {θ_{0}}^{2} \frac{l}{3} = \frac{k {θ_{0}}^{2}}{l}$

Hence the correct answer is $\frac{k {θ_{0}}^{2}}{l} .$

Related content