Question

a simple pendulum is taken to a planet where acceleration due to gravity is 16 times to that of Earth. compare time periods ​

Answer 1

Answer:

Given that, a simple pendulum is taken to a planet where the acceleration due to gravity is 16 times that on earth.

We have to find the time period of the pendulum on the planet.

We know that time period for simple pendulum is:

T = 2π √(l/g) ............(1st equation)

For second's pendulum time period is taken as 2s and acceleration due to gravity is 16 times that on earth.

⇒ acceleration due to gravity (g) = 16g

So,

Let's denote the time period for planet by T' whereas T is the time period for earth.

T' = 2π √(l/16g)

T' = 1/4 × 2π √(l/g) .........(2nd equation)

From (1st equation) we can write the (2md equation) like:

T' = 1/4 × T ........(3rd equation)

As, 2π √(l/g) = T

According to question, we are talking about the second's pendulum and as said above too that time period for second's pendulum is 2s.

Substitute value of T = 2s in (3rd equation)

T' = 1/4 × 2

T' = 2/4

T' = 1/2

T' = 0.5

Therefore, the time period of the pendulum on the planet is 0.5 sec.