Question

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Answer 1

Question :

Find the length of a simple pendulum whose time period of oscillation is 1.8s at a place where acceleration due to gravity is 9.8m/s².

Given :

Time periods = 1.8s

Acceleration due to gravity = 9.8m/s²

To Find :

Length of the pendulum.

Solution :

Time period of simple pendulum in terms of length of pendulum and gravitational acceleration is given by

$\bigstar\:\underline{\boxed{\bf{\orange{T=2\pi\sqrt{\dfrac{L}{g}}}}}}$

• T = 1.8 s
• g = 9.8 m/s²

By substituting the values;

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

$:\dashrightarrow\sf\:1.8=2(3.14)\sqrt{\dfrac{L}{9.8}}$

$:\dashrightarrow\sf\:\dfrac{1.8}{6.28}=\sqrt{\dfrac{L}{9.8}}$

$:\dashrightarrow\sf\:(0.286)^2=\dfrac{L}{9.8}$

$:\dashrightarrow\sf\:L=0.08\times 9.8$

$:\dashrightarrow\:\underline{\boxed{\bf{\purple{L=0.78\:m}}}}$

■ Additional Information :

• The motion of a simple pendulum is simple harmonic for small angular displacement.
• Second pendulum is the simple pendulum, having a time period of 2 second. Its effective length is 99.992 cm or approximate one metre on earth.