Question

The bob of mass $m$ of a simple pendulum, is attached to a horizontal spring of spring constant $k$. The pendulum will undergo simple harmonic motion with period $T$ equal to

• A. $2\pi \sqrt{\frac{L}{g}}$
• B. $2\pi \sqrt{\frac{m}{k}}$
• C. $2\pi \left(\frac{1}{\sqrt{\left(\frac{g}{L}\right)+\left(\frac{k}{m}\right)}}\right)$
• D. $\frac{1}{2}2\pi \sqrt{\left(\frac{L}{g}\right)+\frac{2\pi }{\sqrt{\frac{m}{k}}}}$

Answer 1

The correct option is C $2\pi \left(\frac{1}{\sqrt{\left(\frac{g}{L}\right)+\left(\frac{k}{m}\right)}}\right)$

For small deflection $\theta$, the free body diagram for the bob is drawn.

Total restoring force acting on the bob:
Considering $\theta$ to be small, we can approximate,
$\sin \theta =\theta =\frac{y}{L}$
$F_r=-(mg\sin \theta +kL\theta )F_r=-(mg+kL)\theta$

Acceleration is given as: $a=F_r/m$
$a=-\left(\frac{mg+kL}{m}\right)\frac{y}{L}$
Now we have:
$a\propto -y$, so we can say that bob performs SHM. Comparing above equation with standard equation of SHM, $a=-{\omega }^{2}y$. We get,
$\Rightarrow {\omega }^2=\frac{g}{L}+\frac{k}{m}$

As, $T=\frac{2\pi }{\omega }$
$\therefore T=2\pi \sqrt{\frac{1}{\left(\frac{g}{L}\right)+\left(\frac{k}{m}\right)}}$