Question

A rigid uniform rod AB of length L and mass m is hinged at C such that AC=L/3, CB=3L/3. Ends A and B are supported by springs of spring constant k. The natural frequency of the system is given by

• A. $\sqrt{\frac{k}{2m}}$
• B. $\sqrt{\frac{k}{m}}$
• C. $\sqrt{\frac{2k}{m}}$
• D. $\sqrt{\frac{5k}{m}}$

Answer 1

The correct option is D $\sqrt{\frac{5k}{m}}$

Method I:



$\theta =\frac{y_A}{\frac{L}{3}}=\frac{y_B}{\frac{2L}{3}}$

$I_0=I_C+m{(\frac{2L}{3}-\frac{L}{2})}^2$

$I_C=\frac{m{L}^2}{12}$

$\therefore I_0=\frac{m{L}^2}{12}+\frac{m{L}^2}{36}=\frac{m{L}^2}{9}$

$\text{Taking,}\Sigma M_0=0$

$\Rightarrow (k{y}_A\times \frac{L}{3})+(k{y}_B\times \frac{2L}{3})+I_0\ddot{\theta }=0$

$\Rightarrow k\theta {\left(\frac{L}{3}\right)}^2+k\theta {\left(\frac{2L}{3}\right)}^2+\frac{m{L}^2}{9}\ddot{\theta }=0$

$\Rightarrow \left(\frac{m{L}^2}{9}\right)\ddot{\theta }+\left(\frac{5k{L}^2}{9}\right)\theta =0$

$\Rightarrow \ddot{\theta }+\left(\frac{5k}{m}\right)\theta =0$

$\therefore {\omega }_n^2=\frac{5k}{m};{\omega }_n=\sqrt{\frac{5k}{m}}$

Method II :



Restoring torque:

$T=\left(k\frac{L}{3}\theta \right)\frac{L}{3}+\frac{2L}{3}\left(k\frac{2L}{3}\theta \right)$

$I_c\alpha =\left(\frac{5k{L}^2}{9}\right)\theta$

$\alpha =\frac{\left(\frac{5k{L}^2}{9}\right)\theta }{\frac{mL}{12}+m{\left(\frac{L}{6}\right)}^2}=\left(\frac{5k}{m}\right)\theta$

We know that,

${\omega }_n=\sqrt{\frac{\text{Angular acceleration}}{\text{Angular displacement}}}$

$=\sqrt{\frac{\left(\frac{5k}{m}\right)\theta }{\theta }}=\sqrt{\frac{5k}{m}}$