Question

Compare the time periods of two simple pendulums of length 1 m and 16 m at a
place . (Use T = 2π

)

Answer 1

Answer:

Explanation:

t=2pirootl/g

timeperiod of length 1m=2piroot1/10

timeperiod of length 16m=2piroot16/10

so their ratio of timeperiod is

1:4

Answer 2

Solution :

Given :-

Length of two pendulums are 1m and 16m respectively.

To Find :-

Relation between time period of both pendulums.

Concept :-

Time period is defined as time requires to complete one oscillation.

Formula of time period of simple pendulum in terms of length of pendulum and gravitational acceleration is given by

$\underline{\boxed{\bf{\pink{T=2\pi\sqrt{\dfrac{l}{g}}}}}}$

• T denotes time period
• l denotes length of pendulum
• g denotes gravitational acceleration

Calculation :-

Since, gravitational acceleration is same for both, we can say that...

$\implies\bf\:\red{T\propto \sqrt{l}}\\ \\ \implies\sf\:\dfrac{T_1}{T_2}=\sqrt{\dfrac{l_1}{l_2}}\\ \\ \implies\sf\:\dfrac{T_1}{T_2}=\sqrt{\dfrac{1}{16}}\\ \\ \implies\boxed{\bf{\orange{T_1:T_2=1:4}}}$