Question

A rod of mass M and length L is hinged at its one end and carries a block of mass m at its lower end. A spring of force constant k₁ is installed at distance a from the hinge and another of force constant k₂ at a distance b as shown in the figure. If the whole arrangement rests on a smooth horizontal table top, find the frequency of vibration.

• A. $\frac{1}{2 \pi } \sqrt{\frac{k_{1} a^{2} + k_{2} b^{2}}{L^{2} \left(\right. m + \frac{M}{3} \left.\right)}}$
• B. $\frac{1}{2 \pi } \sqrt{\frac{k_{1} a^{2} + k_{2} b^{2}}{L^{2} \left(\right. m - \frac{M}{3} \left.\right)}}$
• C. $\frac{1}{2 \pi } \sqrt{\frac{k_{1} a^{2} - k_{2} b^{2}}{L^{2} \left(\right. m + \frac{M}{3} \left.\right)}}$
• D. $\frac{1}{4 \pi } \sqrt{\frac{k_{1} a^{2} + k_{2} b^{2}}{L^{2} \left(\right. m - \frac{M}{3} \left.\right)}}$

Answer 1

Answer: (A)

$\tau=\left(\mathrm{k}_1 \mathrm{a} \theta\right) \mathrm{a}+\left(\mathrm{k}_2 \mathrm{~b} \theta\right) \mathrm{b} $
or $ \mathrm{I} \alpha=\left(\mathrm{k}_1 \mathrm{a}^2+\mathrm{k}_2 \mathrm{~b}^2\right) \theta$
or $\left(\mathrm{mL}^2+\frac{1}{3} \mathrm{ML}^2\right) \alpha=\left(\mathrm{k}_1 \mathrm{a}^2+\mathrm{k}_2 \mathrm{~b}^2\right) \theta$
or $ \alpha=\frac{\left(\mathrm{k}_1 \mathrm{a}^2+\mathrm{k}_2 \mathrm{~b}^2\right) \theta}{\mathrm{L}^2\left(\mathrm{~m}+\frac{\mathrm{M}}{3}\right)}$
$\therefore \omega^2=\frac{\mathrm{k}_1 \mathrm{a}^2+\mathrm{k}_2 \mathrm{~b}^2}{\mathrm{~L}^2\left(\mathrm{~m}+\frac{\mathrm{M}}{3}\right)}$

Hence, frequency
$f =\frac{2}{\pi}=\frac{1}{2 \pi} \sqrt{\frac{ k _{1} a ^{2}+ k _{2} b ^{2}}{ L ^{2}\left( m +\frac{ M }{3}\right)}}$