Question

Find the period of small oscillations of the bob of mass $m$ shown in the figure. Given, mass of the rod is also $m$.

• A. $T=2\pi \sqrt{\frac{2m}{3k}}$
• B. $T=2\pi \sqrt{\frac{m}{3k}}$
• C. $T=2\pi \sqrt{\frac{7m}{3k}}$
• D. $T=2\pi \sqrt{\frac{4m}{3k}}$

Answer 1

The correct option is D $T=2\pi \sqrt{\frac{4m}{3k}}$

Suppose the small mass $m$ is pulled down by a small distance $x$ and released.


From conservation of energy principle, we obtain
$\frac{1}{2}k{x}^2-mgx+\frac{1}{2}m{v}^2+\frac{1}{2}l{\omega }^2=\text{constant}.....\left(1\right)$
where $v$ is speed of the mass $m$ when the elongation of the spring is $x$ & $\omega$ is the angular speed of the rod.
Neglecting loss of gravitational P.E term,
$\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2+\frac{1}{2}l{\omega }^2=\text{constant}.....\left(2\right)$
As $x$ is very small, $x=\frac{l\theta }{2}$.
$\Rightarrow v=\frac{dx}{dt}=\frac{l\omega }{2}$
Putting this value in (2) we obtain,
$\frac{1}{2}k{\left(\frac{l}{2}\theta \right)}^2+\frac{1}{2}m{(\frac{l}{2}.\omega )}^2+\frac{1}{2}\frac{m{l}^2}{12}{\omega }^2=\text{constant}$
Differentiating w.r.t time, we obtain
$\frac{1}{2}k\frac{l^2}{4}.2\theta .\frac{d\theta }{dt}+\frac{1}{2}m.\frac{l^2}{4}2\omega \frac{d\omega }{dt}+\frac{1}{2}\frac{m{l}^2}{12}.2\omega \frac{d\omega }{dt}=0$.
$\Rightarrow \frac{d\omega }{dt}=-\frac{k{l}^2}{4}\left[\frac{1}{\frac{m{l}^2}{4}+\frac{m{l}^2}{12}}\right].\theta$
$\Rightarrow \frac{d\omega }{dt}=-\frac{3k}{4m}\theta$
Since $\frac{d\omega }{dt}=\alpha$.
$\Rightarrow \alpha =-\frac{3k}{4m}.\theta$
Compare with $\alpha =-{\omega }_{osc}^2\theta$.
Therefore the angular frequency of oscillation is given by, ${\omega }_{osc}=\sqrt{\frac{3k}{4m}}$
$\Rightarrow T=2\pi \sqrt{\frac{4m}{3k}}$ is the time period of osillations.