Question

Calculate the length of a seconds pendulum at a place where g=9.8 m/s2

Answer 1

Second pendulum is that simple pendulum whose time period of vibrations is 2 seconds.
Since, time period of simple pendulum is given by :
T = 2 π lg−−√
Putting T = 2 s and g = 9.8 m/s2 in above equation, we get :
l = 99.3 cm

Answer 2

Answer:

The length of a seconds pendulum whose time period $T=2s$ is $l=0.99m$

Explanation:

The seconds pendulum is a type of pendulum whose period is exactly two seconds.

$T=2s$

There are only two swings with a one-second duration for each.

Its frequency is $f=1/2=0.5Hz$

We have the time period of a normal pendulum is

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

For a point mass in a weightless string of length $l$ and time period $T=2s$

we can rearrange the equation as

$4=4\pi ^{2} l/g\\l=g/\pi ^{2}$

from the question $g=9.8m/s^{2}$

therefore $l=9.8/\pi ^{2}=9.8/9.85=0.99m$