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, lthe time period changes to 3.99 seconds. Making an approximate analysis, find the acceleration of the car.

Answer

`g/10ms^-2`

Explanation

When the car moving with uniform velocity
`T=2pi sqrt(l/g)`
`rarr 4=2pisqrt(l/g)`……….1
When the car makes accelerated motion let the acceleration be `a_0`.
`T=2pisqrt(l/((g^2+a_0^2)^(1/2)))`
`rarr 3.99=2pi l/(g^2+a_0^2)`
now, `T/T=4/3.99=((g^2+a_0^2)^(1/4))`
Solving for `a_0` we can get
`a_0=g/10ms^-2`