Question

A simple pendulum hangs from the ceiling of a car. Time period of pendulum when the car is at rest was found $2sec$. Now if the car moves uniformly in a large circle, pendulum stays in equilibrium with respect to car at an angle of ${36}^0$ with the vertical. The new time period of the pendulum for small oscillations is : (use $cos{36}^0=0.81$)

• A. $2$ $sec.$
• B. $1.8$ $sec.$
• C. $1.62$ $sec$
• D. $1.4$ $sec.$

Answer 1

The correct option is B $1.8$ $sec.$

Answer is B
We know that time period of simple pendulum is,
$T=2\pi \sqrt{\frac{l}{g}}=2sec\left(i\right)$
Now it makes angle 36 degrees when car is moving in a circle. Therefore component of acceleration due to net force acting on the string will be
$=\sqrt{g^2+a^2}$
New Time period, T', will be
$T^{\prime }=2\pi \sqrt{\frac{l}{\sqrt{g^2+a^2}}}$
$T^{\prime }=2\pi \sqrt{\frac{l}{g\sqrt{1+\frac{a^2}{g^2}}}}\left(ii\right)$
Also,
$tan\theta =a/g$ (iii)
Solving i, ii and iii, we get
$T^{\prime }=2\sqrt{cos\theta }$
$=2\sqrt{0.81}=1.8sec$