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,
$$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{l}{g}} \\ \Rightarrow 4=2\mathrm{\pi }\sqrt{\frac{l}{g}} \end{gathered}$$
Let the acceleration of the car be a.
The time period of pendulum, when the car is accelerated, is given by:
$$\begin{gathered} T\text{'}=2\mathrm{\pi }\sqrt{\frac{l}{{\left(g^2+a^2\right)}^{{\displaystyle \frac{1}{2}}}}} \\ \Rightarrow 3.99=2\mathrm{\pi }\sqrt{\frac{l}{{\left(g^2+a^2\right)}^{\frac{1}{2}}}} \\ \mathrm{Taking}\mathrm{the}\mathrm{ratio}\mathrm{of}T\mathrm{to}T\mathit{\text{'}},\mathrm{we}\mathrm{get}: \\ \frac{T}{T\text{'}}=\frac{4}{3.99}=\frac{{\left(g^2+a^2\right)}^{1/4}}{\sqrt{g}} \end{gathered}$$
On solving the above equation for a, we get:
$a=\frac{g}{10}{\mathrm{ms}}^{-2}$