Question

The lengths of two simple pendulums at a place are in ratio 9:1. Find the ratio of their time periods.

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Answer 1

Answer :

• Ratio of the time periods of the pendulum is 3 : 1.

Explanation :

Given :.

• Ratio of the time periods of the pendulum, $\sf{L_{1} : L_{2} = 9 : 1}$

To find :

• Ratio of the time periods of the pendulum, $\sf{t_{1} : t_{2} = ?}$

Knowledge required :

Formula for time period of a pendulum :

$\boxed{\sf{T = 2\pi\sqrt{\dfrac{L}{g}}}}$

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 :

$:\implies \sf{T = 2\pi\sqrt{\dfrac{L}{g}}} \\ \\ :\implies \sf{T_{1} = 2\pi\sqrt{\dfrac{9}{g}}} \\ \\ \boxed{\therefore \sf{T_{1} = 2\pi\sqrt{\dfrac{9}{g}}}}$

• 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 :

$:\implies \sf{T = 2\pi\sqrt{\dfrac{L}{g}}} \\ \\ :\implies \sf{T_{2} = 2\pi\sqrt{\dfrac{1}{g}}} \\ \\ \boxed{\therefore \sf{T_{2} = 2\pi\sqrt{\dfrac{9}{g}}}}$

Now let's find out the ratio of the time periods.

$:\implies \sf{T_{1} : T_{2} = \dfrac{2\pi\sqrt{\dfrac{9}{g}}}{2\pi\sqrt{\dfrac{1}{g}}}} \\ \\ \\ :\implies \sf{T_{1} : T_{2} = \dfrac{\sqrt{\dfrac{9}{g}}}{\sqrt{\dfrac{1}{g}}}} \\ \\ \\ :\implies \sf{T_{1}^{2} : T_{2}^{2} = \dfrac{\dfrac{9}{g}}{\dfrac{1}{g}}} \\ \\ \\ :\implies \sf{T_{1}^{2} : T_{2}^{2} = \dfrac{9}{g} \times \dfrac{g}{1}} \\ \\ \\ :\implies \sf{T_{1}^{2} : T_{2}^{2} = \dfrac{9}{1}} \\ \\ \\ :\implies \sf{T_{1} : T_{2} = \sqrt{\dfrac{9}{1}}} \\ \\ \\ :\implies \sf{T_{1} : T_{2} = \dfrac{3}{1}} \\ \\ \\ \boxed{\therefore \sf{T_{1} : T_{2} = 3 : 1}}$

Therefore,

• Ratio of the time periods of the pendulum, $\sf{t_{1} : t_{2} = 3 : 1}$

Answer 2

Answer :

$The \: ratio \: of \: their \: time \: periods \: = \: T1 : T2 = 3:1$

To find :

The Ratio of the time periods of the pendulum t1 : t2 =?

Explanation :

T=2pi × sq.root (L/g)

L=gT^2/4pi

L1:L2=9:1

gT1^2/4pi : gT2^2/4pi = 9:1

T1^2 : T2^2 = 9:1

Hence, T1:T2 = 3:1

More to know :

T = Time period

L = Pendulum length

G = Acceleration due to gravity

SQ = Square root

• Pendulum, body suspended from a fixed point so that it can swing back and forth under the influence of gravity.

• The formula for the period T of a pendulum is T = 2π Square root of√L/g, where L is the length of the pendulum and g is the acceleration due to gravity.