Question 89: Oscillations

Question

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :

Accepted Answer

The formula for the period of a simple pendulum is given by:
$$ T = 2\pi\sqrt{\frac{l}{g}} $$
where:
T is the period of the pendulum,
l is the length of the pendulum, and
g is the acceleration due to gravity.
We need to find g. Rearranging the formula for g, we get:
$$ g = \frac{4\pi^2l}{T^2} $$
Given:
l = 2 m,
T = 2 s,
Substituting these values into the equation, we get:
$$ g = \frac{4\pi^2 	imes 2}{(2)^2} = 2\pi^2 \, ms^{-2} $$

Answer

16 m/s
2

Answer

2$$\pi$$
2
ms
$$-$$2

Answer

$$\pi$$
2
ms
$$-$$2

Answer

9.8 ms
$$-$$2

Oscillations

If the time period of a two meter long simple pendulum is 2s, the acceleration due to gravity at the place where pendulum is executing S.H.M. is :

Choose the correct answer: