Question

A simple pendulum of length $100\text{cm}$ is suspended from the ceiling of a cart which is sliding without friction on an inclined plane of inclination ${60}^{\circ }$. Find the time period of the pendulum.
Take $g=10{\text{m/s}}^2$.

• A. $2\text{s}$
• B. $2.8\text{s}$
• C. $5\text{s}$
• D. $2.14\text{s}$

Answer 1

The correct option is B $2.8\text{s}$

Length of simple pendulum, $l=100\text{cm}$
Angle of inclination $\left(\theta \right)={60}^{\circ }$
Acceleration of cart along the inclined plane is,
$a=\frac{mg\sin \theta }{m}=g\sin \theta$
Weight$\left(mg\right)$ can be resolved into two components $mg\sin \theta$ along the plane and $mg\cos \theta$ perpendicular to the plane.
Effective weight of pendulum bob,
$m\overrightarrow{g^{\prime }}=m\overrightarrow{g}+\overrightarrow{F_P}...\left(1\right)$

where $F_P=ma=mg\sin \theta$ acts on bob, in a direction opposite to its acceleration.

After vectorial addition from Eq.$\left(1\right)$,
$\Rightarrow m{g}^{\prime }=mg\cos \theta$
$\therefore g^{\prime }=g\cos \theta$
Hence $g_{\text{eff}}=g\cos \theta$
Time period of pendulum in the cart is given as,
$T=2\pi \sqrt{\frac{l}{g_{eff}}}$
$\Rightarrow T=2\pi \sqrt{\frac{l}{g\cos \theta }}\therefore T=2\times 3.14\times \sqrt{\frac{1}{5}}=2.8\text{s}$