Question

Figure shows a single degree of freedom system. The system consists of a massless rigid bar OP hinged at O and a mass m at end P. The natural frequency of vibration of the system is

• A. $f_n=\frac{1}{2\pi }\sqrt{\frac{k}{4m}}$
• B. $f_n=\frac{1}{2\pi }\sqrt{\frac{k}{2m}}$
• C. $f_n=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$
• D. $f_n=\frac{1}{2\pi }\sqrt{\frac{2k}{m}}$

Answer 1

The correct option is A $f_n=\frac{1}{2\pi }\sqrt{\frac{k}{4m}}$

Method :I



$\therefore$ moving mass m by angle $\theta$

$\therefore \text{Restoring toque}=kx\times a=k\times \left(a\theta \right)\times a$

$\text{Restoring torque}=\text{inertia torque}$

$k\left(a^2\theta \right)=-I\alpha$

$k\left(a^2\theta \right)=-m(2a{)}^2\ddot{\theta }$

$k\theta =-m\times 4\ddot{\theta }$

$4m\ddot{\theta }+k\theta =0$

$\ddot{\theta }+\frac{k}{4m}\theta =0$ $\therefore$

${\omega }_n=\sqrt{\frac{k}{4m}},\frac{2\pi }{T}=\sqrt{\frac{k}{4m}}$

$T=2\pi \sqrt{\frac{4m}{k}}$

$\therefore f_n=\frac{1}{T}=\frac{1}{2\pi }\sqrt{\frac{k}{4m}}$

Method II:
$\therefore$ Rod is rigid, so it will not deform, for the triangle similarity

$\frac{{\delta }_1}{a}=\frac{{\delta }_2}{2a}$

$\frac{{\delta }_1}{{\delta }_2}=\frac{1}{2}$

${\delta }_2=2{\delta }_1$

Taking moment about point O,

$F\times a-mg\times 2a=0$

$F=2mg$

Now
Deflection,
${\delta }_1=\frac{F}{k}=\frac{2mg}{k}$

Deflection:
${\delta }_2=2{\delta }_1=\frac{4mg}{k}$

Natural frequency of vibration of system,

$f_n=\frac{1}{2\pi }\sqrt{\frac{2}{{\delta }_2}}=\frac{1}{2\pi }\sqrt{\frac{2}{\left(4mg\right)k}}$

$=\frac{1}{2\pi }\sqrt{\frac{k}{4m}}$