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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.
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Solution
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:
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