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, $T=2\mathrm{\pi }\sqrt{\frac{l}{g}}\phantom{\rule{0ex}{0ex}}⇒4=2\mathrm{\pi }\sqrt{\frac{l}{g}}$ Let the acceleration of the car be a. The time period of pendulum, when the car is accelerated, is given by: $T\text{'}=2\mathrm{\pi }\sqrt{\frac{l}{{\left({g}^{2}+{a}^{2}\right)}^{\frac{1}{2}}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒3.99=2\mathrm{\pi }\sqrt{\frac{l}{{\left({g}^{2}+{a}^{2}\right)}^{\frac{1}{2}}}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Taking}\mathrm{the}\mathrm{ratio}\mathrm{of}T\mathrm{to}T\mathit{\text{'}},\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}\frac{T}{T\text{'}}=\frac{4}{3.99}=\frac{{\left({g}^{2}+{a}^{2}\right)}^{1/4}}{\sqrt{g}}$ On solving the above equation for a, we get: $a=\frac{g}{10}{\mathrm{ms}}^{-2}$

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