Question

Calculate the time period of a simple pendulum 39.2 cm long at a place where g=9.8 m/s2

Answer 1

Answer:

1.257s

Explanation:

T=2(pi)(l/g)^(1/2)

T=2×3.14×(0.392/9.8)^(1/2)

T=1.257s

Answer 2

Answer:

The time period of a simple pendulum $39.2cm$ long at a place where $g=9.8m/s^{2}$ is $T=1.25s$

Explanation:

The time period of a simple pendulum whose length $l$ is given by the formula,

$T=2\pi \sqrt{\frac{l}{g} }$

where $g=9.8m/s^{2}$ is the acceleration due to gravity

The time period depends only on the length of the pendulum and acceleration due to gravity which is a constant at a place.

From the question, it is given that the length of the simple pendulum is $l=39.2cm=0.392m$

Substituting the given values in the time period formula we get

$T=2\pi \sqrt{\frac{0.392}{9.8} } =1.25s$

so, the time period of the simple pendulum is $T=1.25s$