Question

in a simple pendulum length increases by 4% and G increased by 2% then time period of simple pendulum​

Answer 1

The time period of simple pendulum increase by 1% behalf of percentage error.

Explanation:

As we have , the Simple pendulum formula,

$T = 2\pi\sqrt{\frac{l}{g} }$.

Given Length increase by 4% i.e = 1.04l

and acceleration due to gravity increase by 2% = 1.02g

Then we have

$T^{'} = 2\pi\sqrt{\frac{1.04l}{1.02g} }$

$T^{'} = \sqrt{\frac{1.04}{1.02} } T$

Then by percentage error

Percentage error (T) = $\frac{T^{'}-T }{T}$

$\frac{ΔT}{T}$ = $\frac{\frac{\sqrt{1.04} }{\sqrt{1.02} }T - T }{T}$

$\frac{ΔT}{T} = 0.009$

$\frac{ΔT}{T}x100 = 0.9% = 1%$ = 1%