Question

A pendulum with a period of 2.00000 s in one location (g = 9.80 m/s^{2} 2 ) is moved to a new location where the period is now 1.99796 s. What is the acceleration due to gravity at its new location?

Answer 1

Acceleration due to gravity at new location will be
`g=9.820022m/sec^2`

Step-by-step explanation:

We have given time period of pendulum T = 2 sec

Acceleration due to gravity
`g=9.8m/sec^2`

Time period of pendulum is given by
`T_1=2\pi \sqrt{(L)/(g)}`, here L is length and G is acceleration due to gravity

So
`2=2\pi \sqrt{(L)/(9.8)}`------eqn 1

In second case time period is 1.99796 sec

So
`1.99196=2\pi \sqrt{(L)/(g)}`--------eqn 2

Now dividing eqn 1 by eqn 2

`(2)/(1.99796)=\sqrt{(g)/(9.8)}`

`\sqrt{(g)/(9.8)}=1.00102`

Squaring both side

`(g)/(9.8)=1.00204`

`g=9.820022m/sec^2`