Question

Compare the time periods of two simple pendulums of length 1m and 16

m at a place​

Answer 1

Answer:

Two simple pendulum of length 1m and 16m respectively are both given small displacement in the same direction of the same instant. They will be phase after one shorter pendulum has complated a oscillations

Answer 2

Answer:

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=2π

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...

\begin{gathered}\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}}}\end{gathered}

⟹T∝

l

⟹

T

2

T

1

=

l

2

l

1

⟹

T

2

T

1

=

16

1

⟹

T

1

:T

2

=1:4