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. $\frac{1}{2\pi }\sqrt{\frac{k_1{a}^2+k_2{b}^2}{L^2(m+\frac{M}{3})}}$
• B. $\frac{1}{2\pi }\sqrt{\frac{k_1{a}^2+k_2{b}^2}{L^2(m-\frac{M}{3})}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{k_1{a}^2-k_2{b}^2}{L^2(m+\frac{M}{3})}}$
• D. $\frac{1}{4\pi }\sqrt{\frac{k_1{a}^2+k_2{b}^2}{L^2(m-\frac{M}{3})}}$

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
$f=\frac{2}{\pi }=\frac{1}{2\pi }\sqrt{\frac{k_1{a}^2+k_2{b}^2}{L^2\left(m+\frac{M}{3}\right)}}$