2
24
242
2x
Answers
Answer :
Ratio of the time periods of the pendulum is 3 : 1.
Explanation :
Given :.
Ratio of the time periods of the pendulum, \sf{L_{1} : L_{2} = 9 : 1}L
1
:L
2
=9:1
To find :
Ratio of the time periods of the pendulum, \sf{t_{1} : t_{2} = ?}t
1
:t
2
=?
Knowledge required :
Formula for time period of a pendulum :
\boxed{\sf{T = 2\pi\sqrt{\dfrac{L}{g}}}}
T=2π
g
L
Where,
T = Time period of the pendulum.
2π = Constant
L = Length of the pendulum
g = Acceleration due to gravity
Solution :
Let the length of the two simple pendulum in terms of x as 9x and 1x.
So first let us find the time period of both the pendulums , individually :
Time period of the pendulum with Length of 9x.
By using the formula for time period of a pendulum and substituting the values in it, we get :
\begin{gathered}:\implies \sf{T = 2\pi\sqrt{\dfrac{L}{g}}} \\ \\ :\implies \sf{T_{1} = 2\pi\sqrt{\dfrac{9}{g}}} \\ \\ \boxed{\therefore \sf{T_{1} = 2\pi\sqrt{\dfrac{9}{g}}}} \\ \\ \end{gathered}
:⟹T=2π
g
L
:⟹T
1
=2π
g
9
∴T
1
=2π
g
9
Time period of the pendulum with Length of 1x.
By using the formula for time period of a pendulum and substituting the values in it, we get :
\begin{gathered}:\implies \sf{T = 2\pi\sqrt{\dfrac{L}{g}}} \\ \\ :\implies \sf{T_{2} = 2\pi\sqrt{\dfrac{1}{g}}} \\ \\ \boxed{\therefore \sf{T_{2} = 2\pi\sqrt{\dfrac{9}{g}}}} \\ \\ \end{gathered}
:⟹T=2π
g
L
:⟹T
2
=2π
g
1
∴T
2
=2π
g
9
Now let's find out the ratio of the time periods.
\begin{gathered}:\implies \sf{T_{1} : T_{2} = \dfrac{2\pi\sqrt{\dfrac{9}{g}}}{2\pi\sqrt{\dfrac{1}{g}}}} \\ \\ \\ :\implies \sf{T_{1} : T_{2} = \dfrac{\sqrt{\dfrac{9}{g}}}{\sqrt{\dfrac{1}{g}}}} \\ \\ \\ :\implies \sf{T_{1}^{2} : T_{2}^{2} = \dfrac{\dfrac{9}{g}}{\dfrac{1}{g}}} \\ \\ \\ :\implies \sf{T_{1}^{2} : T_{2}^{2} = \dfrac{9}{g} \times \dfrac{g}{1}} \\ \\ \\ :\implies \sf{T_{1}^{2} : T_{2}^{2} = \dfrac{9}{1}} \\ \\ \\ :\implies \sf{T_{1} : T_{2} = \sqrt{\dfrac{9}{1}}} \\ \\ \\ :\implies \sf{T_{1} : T_{2} = \dfrac{3}{1}} \\ \\ \\ \boxed{\therefore \sf{T_{1} : T_{2} = 3 : 1}} \\ \\ \\ \end{gathered}
:⟹T
1
:T
2
=
2π
g
1
2π
g
9
:⟹T
1
:T
2
=
g
1
g
9
:⟹T
1
2
:T
2
2
=
g
1
g
9
:⟹T
1
2
:T
2
2
=
g
9
×
1
g
:⟹T
1
2
:T
2
2
=
1
9
:⟹T
1
:T
2
=
1
9
:⟹T
1
:T
2
=
1
3
∴T
1
:T
2
=3:1
Therefore,
Ratio of the time periods of the pendulum, \sf{t_{1} : t_{2} = 3 : 1}t
1
:t
2
=3:1
Answer:
2424..................