AB and CD are two parallel chords and O is the centre. If AB=24cm, CD=18cm and radius = 15cm , find the distance between the chords if they are on the opposite sides of the
centre.
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:
In the figure, chords AB∥CD
O is the centre of the circle
Radius of the Circle = 15 cm
Length of AB = 24 cm and CD = 18 cm
Join OA and OC
AM = MB = 24/2 = 12 cm
Similarly ON⊥CD
CN = ND = 18/2 = 9 cm
In right △AMO
OA
2
=OM
2
+AM
2
OM
2
=OA
2
−AM
2
OM
2
=15
2
+12
2
=225−144
OM=
81
=9
Similarly in right △CNO
OC
2
=CN
2
+ON
2
(15)
2
=(9)
2
+ON
2
225=81+ON
2
ON
2
=225−81
ON=12cm
Now MN=OM+ON=9+12=21cm
solution
