Question

A simple pendulum $50\text{cm}$ long is suspended from the roof of a cart accelerating in the horizontal direction with constant acceleration of $\sqrt{3}g{\text{m/s}}^2$. The period of small oscillations of the pendulum about its equilibrium position is -
$(g={\pi }^2{\text{m/s}}^2)$

• A. $1.0\text{sec}$
• B. $\sqrt{2}\text{sec}$
• C. $1.53\text{sec}$
• D. $1.68\text{sec}$

Answer 1

The correct option is A $1.0\text{sec}$

Given:
Length of pendulum, $L=50\text{cm}=0.5\text{m}$
Acceleration of cart, $a=\sqrt{3}g{\text{m/s}}^2$
Acceleration due to gravity, $g={\pi }^2{\text{m/s}}^2$
To find:
Time period of small oscillation of pendulum in cart, $T=?$

With respect to the cart, equilibrium position of the pendulum is shown in the figure.

If displaced by small angle $\delta \theta$ from this position, then it will execute $SHM$ about this equilibrium position, time period of which is given by ,
$T=2\pi \sqrt{\frac{L}{g_{eff}}}$ .........(1)
where, $g_{eff}=\sqrt{g^2+a^2}$
$\Rightarrow g_{eff}=\sqrt{g^2+(\sqrt{3}g{)}^2}=2g$ .........(2)
On putting (2) in (1)
$T=2\pi \sqrt{\frac{L}{2g}}$
$T=2\pi \sqrt{\frac{0.5}{2{\pi }^2}}$
$\Rightarrow T=1.0\text{second}$