Math, asked by niranjananiranjana23, 5 months ago

2
24
242
2x
find \: the \: value \: of \: x \: in \: adjoining \: figure

Answers

Answered by Simrankaur1025
2

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

=

g

1

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

Answered by anwesharoy403
1

Answer:

2424..................

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