Question

The effective length of a pendulum is 400 cm. What is its approximate time period?

Answer 1

Answer :

• Time period of the pendulum, T = 0.942 s

Explanation :

Given :

• Length of the pendulum, l = 400 cm or 4 m
• Acceleration due to gravity, g = 10 m/s²
• Value of π = 22/7

To find :

• Time period of the pendulum, T = ?

Knowlwdge required :

Formula for time period of a pendulum :

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

Where,

• T = Time period of the pendulum
• l = Length of the pendulum
• g = Acceleration due to gravity

Solution :

By using the formula for time period of a pendulum and substituting the values in it, we get :

$:\implies \sf{T = 2\pi \sqrt{\dfrac{\it{l}}{g}}}$

$:\implies \sf{T = 2\pi \times \sqrt{\dfrac{4}{10}}}$

$:\implies \sf{T = 2\pi \times \sqrt{0.4}}$

$:\implies \sf{T = 2\pi \times 0.6(approx.)}$

$:\implies \sf{T = 2 \times \dfrac{22}{7} \times 0.6}$

$:\implies \sf{T = \dfrac{22}{7} \times 0.3}$

$:\implies \sf{T = \dfrac{6.6}{7}}$

$:\implies \sf{T = 0.942}$

$\boxed{\therefore \sf{T = 0.942}}$

Hence the time period of the pendulum is 0.942 s.

Answer 2

Explanation:

Answer :

Time period of the pendulum, T = 0.942 s

Explanation :

Given :

Length of the pendulum, l = 400 cm or 4 m

Acceleration due to gravity, g = 10 m/s²

Value of π = 22/7

To find :

Time period of the pendulum, T = ?

Knowlwdge required :

Formula for time period of a pendulum :

\boxed{\sf{T = 2\pi \sqrt{\dfrac{\it{l}}{g}}}}

T=2π

g

l

Where,

T = Time period of the pendulum

l = Length of the pendulum

g = Acceleration due to gravity

Solution :

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{\it{l}}{g}}} \\ \end{gathered}

:⟹T=2π

g

l

\begin{gathered}:\implies \sf{T = 2\pi \times \sqrt{\dfrac{4}{10}}} \\ \end{gathered}

:⟹T=2π×

10

4

\begin{gathered}:\implies \sf{T = 2\pi \times \sqrt{0.4}} \\ \end{gathered}

:⟹T=2π×

0.4

\begin{gathered}:\implies \sf{T = 2\pi \times 0.6(approx.)} \\ \end{gathered}

:⟹T=2π×0.6(approx.)

\begin{gathered}:\implies \sf{T = 2 \times \dfrac{22}{7} \times 0.6} \\ \end{gathered}

:⟹T=2×

7

22

×0.6

\begin{gathered}:\implies \sf{T = \dfrac{22}{7} \times 0.3} \\ \end{gathered}

:⟹T=

7

22

×0.3

\begin{gathered}:\implies \sf{T = \dfrac{6.6}{7}} \\ \end{gathered}

:⟹T=

7

6.6

\begin{gathered}:\implies \sf{T = 0.942} \\ \end{gathered}

:⟹T=0.942

\begin{gathered}\boxed{\therefore \sf{T = 0.942}} \\ \end{gathered}

∴T=0.942

Hence the time period of the pendulum is 0.942 s.