Question

A seconds pendulum is taken to a height where the value of g is 4.36 m/s^2. What will be its new time period

Answer 1

time period= 2π√l/g

it is 1.414

Answer 2

The new time period of the pendulum is 3 seconds.

Explanation:

Given that,

Time period of the seconds pendulum, T = 2 s

Value of acceleration due to gravity on the Earth, $g=9.8\ m/s^2$

The time period of the pendulum is given by :

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

$2=2\pi\sqrt{\dfrac{l}{9.8}}$.............(1)

Let T' s the time period of the simple pendulum when value of g is $4.36\ m/s^2$. New time period is given by :

$T'=2\pi\sqrt{\dfrac{l}{g'}}$............(2)

Dividing equation (1) and (2) we get :

$\dfrac{2}{T'}=\sqrt{\dfrac{4.36}{9.8}}$

T' = 2.99 seconds

or

T' = 3 seconds

So, the new time period of the pendulum is 3 seconds. Hence, this is the required solution.