Question

A rod of mass M and length L is hinged at its center of mass so that it can rotate to a vertical plane.Two springs each of stiffness k are connected at its ends, as shown in the figure, The time period of SHM IS.

• A. $2\pi \sqrt{\frac{M}{6k}}$
• B. $2\pi \sqrt{\frac{M}{3k}}$
• C. $2\pi \sqrt{\frac{Ml}{k}}$
• D. $\pi \sqrt{\frac{M}{6k}}$

Answer 1

The correct option is A $2\pi \sqrt{\frac{M}{6k}}$

If the rod is rotated through an angle θ, extension in one spring = compression in the other spring,

i.e.,$x=l\theta /2$

Therefore, force acting on each of the ends of the rod,

$F=kx=k\left(l\theta /2\right)$

Restoring torque on the rod,

$T=-Fl=-k\left(\theta /2\right)l=-k{l}^2\left(\theta /2\right)$

As $T=I\alpha =\left(M{l}^2/12\right)\alpha$

or $\alpha =-\frac{6K}{M}\theta ,\alpha isproportionalto,\theta$

Thus, the motion of the rod is simple harmonic with

${\omega }^2=\frac{6K}{M}$ $\Rightarrow$ ω = $\sqrt{\frac{6K}{M}}$

$\Rightarrow$ $T$ = $\frac{2\pi }{w}$

$\Rightarrow$ $T$ = $2\pi \sqrt{\frac{M}{6K}}$

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