Question

A simple pendulum fixed in a car has a time period of 4 seconds when the car is moving uniformly on a horizontal road. When the accelerator is pressed, the time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of the car.

Answer 1

It is given that:
When the car is moving uniformly, time period of simple pendulum, T = 4.0 s
As the accelerator is pressed, new time period of the pendulum, T' = 3.99 s
Time period of simple pendulum, when the car is moving uniformly on a horizontal road is given by,

\[T = 2\pi\sqrt{\frac{l}{g}}\] 

\[ \Rightarrow 4 = 2\pi\sqrt{\frac{l}{g}}\]

Let the acceleration of the car be a.
The time period of pendulum, when the car is accelerated, is given by:

\[T' = 2\pi\sqrt{\frac{l}{\left( g^2 + a^2 \right)^\frac{1}{2}}}\] 

\[ \Rightarrow 3 . 99 = 2\pi\sqrt{\frac{l}{\left( g^2 + a^2 \right)^\frac{1}{2}}}\] 

\[\text { Taking  the  ratio  of  T  to  T',   we  get: } \] \[\frac{T}{T'} = \frac{4}{3 . 99} = \frac{\left( g^2 + a^2 \right)^{1/4}}{\sqrt{g}}\]

On solving the above equation for a, we get:

\[a = \frac{g}{10}   {ms}^{- 2}\]